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Polynomials, Exponentials, and Logarithms: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 128 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. A pair of polynomials put through both operations . 11 points. Question 1 of 10.

Let $P = 4x^{3} - x^{2} + 6x - 2$ and $Q = 4x^{3} + 5x^{2} - 6x$ . Write every answer in standard form, with the powers descending.
1. Part A.
  Compute $P + Q$ , and state the degree of the result.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Compute $P - Q$ , and state the degree of the result.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Compare the degrees of your answers to parts A and B. Explain what feature of $P$ and $Q$ decides whether a difference can come out with a lower degree than either polynomial it was built from, and state that condition in general.
  
  Carry your own answer forward Compare whichever two results you produced in parts A and B, even if they were not the expected ones, and argue from what you actually obtained. The credit here is for the account of when a leading term can vanish, not for a particular pair of polynomials.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer
Part A

$P + Q = 8x^{3} + 4x^{2} - 2$ , of degree $3$ .
- $8x^{3} + 4x^{2} + 0x - 2$ is the same polynomial with the empty column written in; a term whose coefficient is $0$ contributes nothing and is normally left out
Part B

$P - Q = -6x^{2} + 12x - 2$ , of degree $2$ .
- $-2 + 12x - 6x^{2}$ is the same polynomial out of order; standard form puts the highest power first, which is where the degree is read
Part C

The sum keeps degree $3$ : the leading coefficients $4$ and $4$ add to $8$ , which is not zero. The difference loses it: they subtract to $0$ . In general the degree can fall below both only when the two polynomials share a degree and the coefficients at that power cancel; otherwise it is the larger of the two.
Worked solution

Part A

A plus sign in front of a parenthesis changes nothing inside it, so both sets of parentheses drop and every sign stays as it stands. Gather the like powers into columns.

$\left(4x^{3} + 4x^{3}\right) + \left(-x^{2} + 5x^{2}\right) + \left(6x - 6x\right) - 2 = 8x^{3} + 4x^{2} - 2$

The $x$ column comes out empty, since $6 - 6 = 0$ , so no $x$ term appears in the answer. The highest power present is $x^{3}$ , so the degree is $3$ .

Part B

The minus sign in front of $Q$ is a factor of $-1$ on the whole group, so all three of its signs turn over.

$4x^{3} - x^{2} + 6x - 2 - 4x^{3} - 5x^{2} + 6x = -6x^{2} + 12x - 2$

Watch the two signs that changed: the $+5x^{2}$ became $-5x^{2}$ and the $-6x$ became $+6x$ . The two $x^{3}$ terms are now equal and opposite, so they wipe each other out, and the highest power left is $x^{2}$ : the degree is $2$ .

Part C

Both $P$ and $Q$ have degree $3$ , and both carry the identical leading term $4x^{3}$ . That coincidence is what the whole question turns on, because the two operations do different things to it.

$4 + 4 = 8 \ne 0, \qquad 4 - 4 = 0$

Adding puts $8x^{3}$ at the front, so the sum keeps degree $3$ . Subtracting puts $0x^{3}$ there, so that power disappears entirely and the degree drops to $2$ , lower than either polynomial started with.

In general, the degree of a sum or difference is at most the larger of the two degrees, and it equals that larger degree unless the two polynomials share it and the coefficients at that power cancel. When the degrees are different, nothing can happen to the higher one at all, because the other polynomial has no term at that power to combine with it.

In one line

$P + Q = 8x^{3} + 4x^{2} - 2$ , of degree $3$ , and $P - Q = -6x^{2} + 12x - 2$ , of degree $2$ . Both polynomials carry the leading term $4x^{3}$ , and $4 + 4 = 8$ keeps it in the sum while $4 - 4 = 0$ wipes it out of the difference. A sum or difference can fall below both original degrees only when the two polynomials share a degree and the coefficients at that power cancel; otherwise the degree is the larger of the two.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Combines only like powers, adding their coefficients and leaving the power itself unchanged. . Worth 2 points.

Writes the result in descending order and names its degree from the highest power actually present. . Worth 1 point.

Part B 4 points

Distributes the minus sign to every term of the second polynomial, not to the first one only. . Worth 2 points.

Combines the like terms correctly and writes the result in descending order. . Worth 1 point.

Names the degree from the highest power that actually survives, not from the polynomials it came from. . Worth 1 point.

Part C 4 points

Locates the cause in the two leading coefficients, showing that one combination is zero and the other is not, rather than quoting a rule. . Worth 3 points. needs an explanation, not just an answer

States the general condition with both of its halves: the same degree, and coefficients at that power that cancel. . Worth 1 point.
2. Exponents read off, and three of them put together . 12 points. Question 2 of 10.

Throughout this question, $\log$ written with no base means base $10$ .
1. Part A.
  Evaluate $\log_{2}(64)$ , $\log_{7}(1)$ and $\log_{5}\!\left(\tfrac{1}{25}\right)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  For some base $b$ you are given $\log_{b}(2) = 0.36$ and $\log_{b}(5) = 0.83$ . Find $\log_{b}(40)$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  The line $\log_{b}(7) = \log_{b}(2) + \log_{b}(5)$ turns up in a page of working, offered on the grounds that $2 + 5 = 7$ . Decide whether it is true, and say precisely which law it is reaching for and what that law requires of its two inputs.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer
Part A

$\log_{2}(64) = 6$ , $\log_{7}(1) = 0$ , and $\log_{5}\!\left(\tfrac{1}{25}\right) = -2$ .
- $6$ , $0$ and $-2$ in that order; what is not the same is $\tfrac{1}{2}$ for the last one, which turns the reciprocal in the input into a reciprocal in the answer instead of a negative exponent
Part B

$\log_{b}(40) = 1.91$ .
- $3(0.36) + 0.83$ is the same value before the arithmetic is finished; $0.36 + 0.83 = 1.19$ is not, because factoring $40$ as $2 \times 5$ loses two whole factors of $2$
Part C

False. The product law turns a product of inputs into a sum of logarithms, so $\log_{b}(2) + \log_{b}(5) = \log_{b}(10)$ , not $\log_{b}(7)$ . The given values show it: the sum is $1.19$ , which is $\log_{b}(10)$ . There is no law at all for the logarithm of a sum.
Worked solution

Part A

Every logarithm asks one question: the base to what power gives the input? Answer it by writing each one in exponential form.

$2^{6} = 64, \qquad 7^{0} = 1, \qquad 5^{-2} = \frac{1}{5^{2}} = \frac{1}{25}$

So the three values are $6$ , $0$ and $-2$ . Any allowed base to the zero power is $1$ , and landing below $1$ takes a negative exponent, not a fractional one.

Part B

Write the input using only the numbers whose logarithms you were handed.

$40 = 8 \times 5 = 2^{3}\cdot 5$

Now take $\log_{b}$ of both sides. The product law splits the two factors into a sum, and the power law pulls the exponent $3$ down in front:

$\log_{b}(40) = \log_{b}\!\left(2^{3}\right) + \log_{b}(5) = 3\log_{b}(2) + \log_{b}(5) = 3(0.36) + 0.83 = 1.91$

The base was never named and was never needed.

Part C

The line is false. The law it is reaching for is the product law,

$\log_{b}(MN) = \log_{b}(M) + \log_{b}(N),$

and what that law requires of its two inputs is that they be MULTIPLIED, not added. Here $2 \times 5 = 10$ , so the right-hand side is the logarithm of $10$ :

$\log_{b}(2) + \log_{b}(5) = 0.36 + 0.83 = 1.19 = \log_{b}(10)$

and $\log_{b}(7)$ is a different number, because a logarithm is one-to-one and $7 \ne 10$ . A small case makes the same point with numbers you can check by hand: with base $2$ , $\log_{2}(4 + 4) = \log_{2}(8) = 3$ , while $\log_{2}(4) + \log_{2}(4) = 2 + 2 = 4$ . Nothing breaks up the logarithm of a sum.

In one line

$\log_{2}(64) = 6$ , $\log_{7}(1) = 0$ and $\log_{5}\!\left(\tfrac{1}{25}\right) = -2$ . Since $40 = 2^{3}\cdot 5$ , the product and power laws give $\log_{b}(40) = 3(0.36) + 0.83 = 1.91$ . The line $\log_{b}(7) = \log_{b}(2) + \log_{b}(5)$ is false: the product law adds logarithms when the inputs are multiplied, so that sum is $\log_{b}(10) = 1.19$ , and there is no law for the logarithm of a sum.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Rewrites each logarithm as the exponential statement it stands for, rather than reaching for a formula. . Worth 2 points.

Reports each answer as the exponent itself, including the negative value for the reciprocal input. . Worth 1 point.

Part B 5 points

Factors the input into powers of the numbers whose logarithms are given, instead of looking for a rule that acts on $40$ directly. . Worth 2 points.

Applies the product law to the two factors and the power law to the exponent, then substitutes the given values. . Worth 2 points.

Reports a single number as the value of the logarithm. . Worth 1 point.

Part C 4 points

Reaches a verdict and grounds it in what the product law actually asks of its two inputs, rather than in the line looking unfamiliar. . Worth 3 points. needs an explanation, not just an answer

Backs the verdict with a computation, either the given values or a small case checked by hand. . Worth 1 point.
3. Every term meeting every term, twice over . 12 points. Question 3 of 10.

Expand each product below and write the answer in standard form.
1. Part A.
  Expand $(3x - 4)(2x + 5)$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Expand $(x + 2)\left(3x^{2} - x + 6\right)$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Parts A and B each asked you to form every product of a term from the first factor with a term from the second. State the rule that decides how many such products a multiplication owes before any like terms are collected. Then explain why the degree of a product can never come out lower than the degree of either factor.
  
  Carry your own answer forward Count the products in whichever expansions you actually carried out in parts A and B. The credit here is for the counting rule and for the argument about the leading terms, not for the two particular answers.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer
Part A

$6x^{2} + 7x - 20$ .
- $6x^{2} + 15x - 8x - 20$ is the same expression with the like terms not yet collected; standard form gathers them into $7x$
Part B

$3x^{3} + 5x^{2} + 4x + 12$ .
- $3x^{3} - x^{2} + 6x + 6x^{2} - 2x + 12$ is the same expression with the six products not yet gathered; standard form combines them
Part C

A factor with $p$ terms times a factor with $q$ terms owes $pq$ products: $4$ in part A and $6$ in part B. The degree cannot drop because the two leading terms multiply into a single term whose degree is the sum of the two factors' degrees; no other pairing reaches that power, and two nonzero coefficients cannot multiply to zero.
Worked solution

Part A

Two terms times two terms owes four products, and every sign travels with its own term.

$(3x)(2x) + (3x)(5) + (-4)(2x) + (-4)(5) = 6x^{2} + 15x - 8x - 20$

The two middle products carry the same power of $x$ , so they combine to $7x$ , and the answer is $6x^{2} + 7x - 20$ .

Part B

Two terms times three terms owes $2 \times 3 = 6$ products, so the four-product pattern is not enough here. Distribute $x$ across the trinomial, then $2$ across it.

$x\left(3x^{2} - x + 6\right) = 3x^{3} - x^{2} + 6x, \qquad 2\left(3x^{2} - x + 6\right) = 6x^{2} - 2x + 12$

Add the two rows and gather like powers: $-x^{2} + 6x^{2} = 5x^{2}$ and $6x - 2x = 4x$ , so the product is $3x^{3} + 5x^{2} + 4x + 12$ .

Part C

How many products. A factor with $m$ terms times a factor with $n$ terms produces $m \times n$ products, one for each pairing, before anything is collected. Part A was $2 \times 2 = 4$ and part B was $2 \times 3 = 6$ . That is exactly why the four-product mnemonic describes a binomial times a binomial and nothing wider.

Why the degree cannot drop. Write each factor in standard form, so its first term is its leading term, say $a x^{m}$ and $b x^{n}$ . Multiply those two:

$\left(a x^{m}\right)\left(b x^{n}\right) = ab\,x^{m+n}.$

Every other product pairs a term of degree at most $m$ with a term of degree at most $n$ , and in each such pairing at least one factor has a strictly smaller degree, so every other product has degree strictly less than $m + n$ . That leaves $ab\,x^{m+n}$ alone at its power with nothing to combine with, and $ab \ne 0$ because neither leading coefficient is zero. So the top term survives and the degree is $m + n$ , which is at least $m$ and at least $n$ .

A sum behaves differently for one reason only: there the two leading terms can land in the same column and be ADDED, and two nonzero numbers can add to zero. In a product they are multiplied, and nonzero times nonzero is never zero.

In one line

$(3x - 4)(2x + 5) = 6x^{2} + 7x - 20$ and $(x + 2)\left(3x^{2} - x + 6\right) = 3x^{3} + 5x^{2} + 4x + 12$ . A factor with $p$ terms times a factor with $q$ terms owes $pq$ products, which is $4$ and then $6$ here. The degree of a product is always the sum of the two factors' degrees: the leading terms multiply into a single term at that power, no other pairing reaches it, and a product of two nonzero coefficients is nonzero, so it cannot be cancelled. A sum can lose its top term only because addition, unlike multiplication, can turn two nonzero numbers into zero.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Forms all four products, carrying each sign into the product it belongs to. . Worth 2 points.

Collects the two like middle terms and reports the result in descending order. . Worth 1 point.

Part B 4 points

Multiplies every term of the binomial by every term of the trinomial, six products in all rather than four. . Worth 2 points.

Combines the like powers correctly and reports the result in descending order. . Worth 2 points.

Part C 5 points

States the counting rule as the product of the two term counts and matches it against both expansions. . Worth 1 point.

Argues that the two leading terms multiply into a single top-degree term that no other pairing reaches, and that its coefficient cannot be zero. . Worth 3 points. needs an explanation, not just an answer

Runs the argument in general letters, so it covers any two polynomials rather than only the two expanded above. . Worth 1 point.
4. Four outputs, and the curve they belong to . 13 points. Question 4 of 10.

A function $f$ is known to have the form $f(x) = a\cdot b^{x}$ with $a > 0$ , $b > 0$ and $b \ne 1$ . Its outputs at four equally spaced inputs are: at $x = 0$ the output is $7$ , at $x = 1$ it is $4.2$ , at $x = 2$ it is $2.52$ , and at $x = 3$ it is $1.512$ .
1. Part A.
  Show that these outputs are consistent with a rule of that form, and write down $f$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Say whether $f$ rises or falls as $x$ increases, give the value at which its graph meets the vertical axis, state the range of $f$ , and say what value its outputs creep toward as $x$ increases.
  
  Carry your own answer forward Describe whichever rule you wrote down in part A, even if it was not the expected one, and read its features off honestly. The credit here is for reading direction, intercept and range off a rule of the form $a\cdot b^{x}$ , not for a particular base.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 5 points
3. Part C.
  Decide whether $f(x)$ must eventually reach $0$ and then turn negative, and argue from the form of the rule rather than from further values of it.
  
  Carry your own answer forward Run the argument on whichever rule you wrote down in part A, even if it was not the expected one, and say honestly what it does and does not rule out. The credit here is for the reasoning about a positive base raised to a real power, not for a particular number.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer
Part A

The ratios between consecutive outputs are all $0.6$ , so $f(x) = 7\,(0.6)^{x}$ .
- $f(x) = 7\left(\tfrac{3}{5}\right)^{x}$ is the same rule with the base written as a fraction; what is not the same is $f(x) = 7 - 2.8x$ , which matches the first two outputs and then fails at the third
Part B

It falls, since the base $0.6$ lies between $0$ and $1$ . The graph meets the vertical axis at $7$ , the range is $y > 0$ , and the outputs creep toward $0$ as $x$ increases.

Part C

It cannot. For every real $x$ , $0.6^{x}$ is a positive base raised to a real power and so is positive, and $7$ is positive, so every output is a positive times a positive. The outputs shrink toward $0$ without limit but never arrive, which is what the horizontal asymptote $y = 0$ records.
Worked solution

Part A

The inputs step by $1$ , which is the hypothesis that lets a ratio test mean anything. Test the differences first, the linear signature: $4.2 - 7 = -2.8$ but $2.52 - 4.2 = -1.68$ , so the data is not linear. Now test the ratios.

$\frac{4.2}{7} = 0.6, \qquad \frac{2.52}{4.2} = 0.6, \qquad \frac{1.512}{2.52} = 0.6$

The ratio is a constant $0.6$ , and with steps of $1$ that constant ratio is the base. The output at $x = 0$ is the multiplier, so $a = 7$ and $f(x) = 7\,(0.6)^{x}$ . Check the far end: $7 \times 0.6^{3} = 7 \times 0.216 = 1.512$ .

Part B

The base decides the direction. Here $0 < 0.6 < 1$ , so each step to the right multiplies the output by less than $1$ and the graph falls. The vertical intercept is the value at $x = 0$ , where the power is $1$ and only the multiplier is left.

$f(0) = 7\,(0.6)^{0} = 7 \times 1 = 7$

A positive multiplier times a positive base raised to any real power is positive, so every output is positive and the range is $y > 0$ . As $x$ increases, the repeated multiplication by $0.6$ drives the outputs down toward $0$ without their arriving, which is the horizontal asymptote $y = 0$ .

Part C

It never happens, and the reason is in the rule rather than in any table of values. A positive base raised to a real power is positive: a whole-number exponent multiplies positive numbers together, a negative exponent takes the reciprocal of a positive number, and a fractional exponent takes a root of a positive number.

$0.6^{x} > 0 \quad \text{for every real } x, \qquad \text{so} \qquad f(x) = 7\,(0.6)^{x} > 0.$

Reaching $0$ would need $0.6^{x} = 0$ , and no exponent does that: multiplying by $0.6$ over and over makes a positive number small, never zero. Turning negative would need the curve to pass through $0$ on the way, so that is ruled out as well. What the falling outputs do instead is creep toward the horizontal asymptote $y = 0$ , getting arbitrarily close to it and staying above it.

In one line

The ratios $\tfrac{4.2}{7}$ , $\tfrac{2.52}{4.2}$ and $\tfrac{1.512}{2.52}$ are all $0.6$ over equally spaced inputs, so $f(x) = 7\,(0.6)^{x}$ . Because $0 < 0.6 < 1$ the graph falls; it meets the vertical axis at $7$ , its range is $y > 0$ , and its outputs creep toward $0$ . They never reach $0$ and never turn negative, because a positive base raised to any real power is positive and $7$ is positive, so every output is a product of two positive numbers.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Tests the ratios between consecutive outputs, using that the inputs are equally spaced rather than assuming it. . Worth 2 points.

Takes the base from the constant ratio and the multiplier from the output at $x = 0$ , and writes the rule out. . Worth 2 points.

Part B 5 points

Decides the direction from where the base sits relative to $1$ , not from the look of the listed numbers. . Worth 2 points.

Evaluates the rule at $x = 0$ to get the vertical intercept. . Worth 1 point.

States the range and the value the outputs creep toward, and keeps the two consistent with each other. . Worth 2 points.

Part C 4 points

Argues from a positive base raised to a real power always being positive, rather than from the outputs merely looking small. . Worth 3 points. needs an explanation, not just an answer

Describes what the outputs do instead, in terms of the horizontal asymptote, consistently with the sign argument. . Worth 1 point.
5. One rate, three schedules . 13 points. Question 5 of 10.

A deposit of $4000$ dollars earns $5\%$ per year. Round only at the end, and give each balance to the nearest cent. These powers are available, and not all of them are needed: $1.05^{3} = 1.157625$ , $1.05^{12} = 1.795856$ , $1.0125^{3} = 1.037971$ , $1.0125^{12} = 1.160755$ , and $e^{0.15} = 1.161834$ .
1. Part A.
  Find the balance after $3$ years with the interest compounded annually, and again with it compounded quarterly.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Find the balance after $3$ years with the interest compounded continuously, and say how much more it is than the quarterly balance.
  
  Carry your own answer forward Subtract whichever quarterly balance you produced in part A, even if it was not the expected one. The credit here is for using the continuous formula correctly and for comparing it honestly against your own earlier figure.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Compare the three balances you have found. Then explain why raising the compounding frequency again and again cannot make the balance as large as you please, and say what the number $e$ has to do with the limit.
  
  Carry your own answer forward Compare whichever three balances you produced in parts A and B. The credit here is for what the pattern of gains shows and for the account of the ceiling, not for a particular pair of differences.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 5 points
The answer
Part A

$4630.50$ dollars compounded annually, and $4643.02$ dollars compounded quarterly.
- $4630.50$ may be written $4630.5$ ; what is not the same is $4600$ dollars from simple interest, which adds a flat $200$ dollars a year and never pays interest on interest
Part B

$4647.34$ dollars, which is $4.32$ dollars more than the quarterly balance.

Part C

Each rise in frequency earns more, but by less each time: annual to quarterly gains about $12.52$ dollars, quarterly to continuous only about $4.32$ dollars. The factor $\left(1 + \tfrac{r}{n}\right)^{n}$ climbs toward $e^{r}$ rather than without bound, so every finite schedule stays under the ceiling $Pe^{rt}$ .
Worked solution

Part A

The rate is the decimal form of the percent, so $r = 0.05$ . Annual compounding applies the factor $1 + r$ once per year.

$A = 4000\,(1.05)^{3} = 4000 \times 1.157625 = 4630.50$

Quarterly compounding makes two edits, and both are needed: each period pays $\tfrac{0.05}{4} = 0.0125$ , and over $3$ years there are $nt = 4 \times 3 = 12$ periods, not $3$ .

$A = 4000\left(1 + \frac{0.05}{4}\right)^{12} = 4000\,(1.0125)^{12} = 4000 \times 1.160755 = 4643.02$

Part B

Continuous compounding is the limiting case, with $A = Pe^{rt}$ and the whole product $rt = 0.05 \times 3 = 0.15$ in the exponent.

$A = 4000\,e^{0.15} = 4000 \times 1.161834 = 4647.34$

Against the quarterly figure, $4647.34 - 4643.02 = 4.32$ , so the continuous schedule is worth another $4.32$ dollars across the three years.

Part C

The three balances line up as $4630.50 < 4643.02 < 4647.34$ , and the gains between them shrink: $12.52$ dollars from annual to quarterly, then only $4.32$ dollars from quarterly to continuous.

That pattern is not an accident of these numbers. Compounding $n$ times a year pays $\tfrac{r}{n}$ per period over $nt$ periods, and the growth factor regroups as

$\left(1 + \frac{r}{n}\right)^{nt} = \left[\left(1 + \frac{r}{n}\right)^{n}\right]^{t}.$

As $n$ grows, the inner bracket does not run off to infinity. It rises by smaller and smaller amounts toward $e^{r}$ , because $\left(1 + \tfrac{1}{m}\right)^{m}$ crowds toward $e \approx 2.71828$ as $m$ grows. So the whole factor approaches $\left(e^{r}\right)^{t} = e^{rt}$ , and every finite schedule stays below $Pe^{rt}$ . Compounding infinitely often gives a finite answer, and that answer is the continuous balance the other two are climbing toward.

In one line

On $4000$ dollars at $5\%$ for $3$ years the balances are $4630.50$ dollars compounded annually, $4643.02$ dollars compounded quarterly, and $4647.34$ dollars compounded continuously, the last being $4.32$ dollars above the quarterly figure. Each rise in frequency earns less than the rise before it, because $\left(1 + \tfrac{r}{n}\right)^{n}$ climbs toward $e^{r}$ rather than without bound. Every finite schedule therefore stays below the continuous ceiling $A = Pe^{rt}$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Writes the rate as a decimal and, for the quarterly schedule, divides it by $4$ while multiplying the exponent by $4$ . . Worth 2 points.

Carries out both calculations and rounds only the final balances, not the growth factors. . Worth 1 point.

States both answers as amounts of money to the nearest cent. . Worth 1 point.

Part B 4 points

Uses $A = Pe^{rt}$ with the product of rate and time in the exponent, not the rate alone. . Worth 2 points.

Computes the balance and then the gap against the earlier figure. . Worth 1 point.

Reports both the balance and the gap as amounts of money. . Worth 1 point.

Part C 5 points

Orders the three balances and observes that the gains between them are shrinking, not merely that they are positive. . Worth 2 points.

Ties the ceiling to the factor for one year approaching $e^{r}$ , rather than asserting without reason that the balance is bounded. . Worth 2 points. needs an explanation, not just an answer

Names the continuous formula as the ceiling itself, so the limit is a specific amount rather than a vague bound. . Worth 1 point.
6. Two operations, two rules about degree . 12 points. Question 6 of 10.

Let $A = 2x^{3} + x^{2} - 5$ and $B = -2x^{3} + 4x + 1$ .
1. Part A.
  Compute $A + B$ and state its degree.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Without expanding the whole product, state the degree of $A\cdot B$ and its leading term.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
3. Part C.
  Two rules are in play here: that the degree of a sum is the larger of the two degrees, and that the degree of a product is the sum of the two degrees. For each, decide whether it holds for every pair of polynomials or whether it can fail, using your own parts A and B as evidence, and explain what makes the two cases behave differently.
  
  Carry your own answer forward Test both rules against whichever results you produced in parts A and B, even if they were not the expected ones, and report honestly which of your own answers supports or contradicts each rule.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 5 points
The answer
Part A

$A + B = x^{2} + 4x - 4$ , of degree $2$ .
- $0x^{3} + x^{2} + 4x - 4$ is the same polynomial with the empty column written in; a term whose coefficient is $0$ contributes nothing and is normally left out
Part B

$A\cdot B$ has degree $6$ , with leading term $-4x^{6}$ .

Part C

The product rule always holds: the leading terms multiply, and two nonzero coefficients cannot multiply to zero. The sum rule can fail, and part A is a counterexample: $2x^{3}$ and $-2x^{3}$ added to zero, dropping the degree from $3$ to $2$ . Addition can cancel a leading coefficient; multiplication cannot.
Worked solution

Part A

Adding changes no signs, so both sets of parentheses drop and the like powers are gathered as they stand.

$\left(2x^{3} - 2x^{3}\right) + x^{2} + 4x + (-5 + 1) = x^{2} + 4x - 4$

The two leading coefficients are $2$ and $-2$ , which add to $0$ , so the $x^{3}$ column empties and the highest power left is $x^{2}$ . The degree is $2$ .

Part B

Only the two leading terms can reach the top power, so the front of the product is settled without touching the other eight products.

$\left(2x^{3}\right)\left(-2x^{3}\right) = -4x^{6}$

Every other pairing multiplies a term of degree at most $3$ by a term of degree at most $3$ with at least one of them strictly smaller, so no other product reaches degree $6$ . Since $-4$ is not zero, that term survives with nothing at its power to cancel it, and the degree is $3 + 3 = 6$ .

Part C

The product rule holds without exception. The leading terms of $A$ and $B$ are multiplied, not added, and their coefficients are $2$ and $-2$ :

$2 \times (-2) = -4 \ne 0.$

A product of two nonzero numbers is never zero, and no other pairing of terms reaches degree $6$ , so the top term of $A\cdot B$ always survives. That is why the degree of a product is the sum of the degrees, with no exceptions.

The sum rule can fail, and part A shows it failing. There the same two leading coefficients meet under addition instead:

$2 + (-2) = 0,$

so the $x^{3}$ term disappears and $A + B$ has degree $2$ , not the $3$ the rule as stated would predict. The honest version is that the degree of a sum is at most the larger of the two degrees, and equals it unless the two polynomials share that degree and their leading coefficients cancel.

The whole difference is which operation the two leading coefficients meet under. Addition can send two nonzero numbers to zero; multiplication cannot.

In one line

$A + B = x^{2} + 4x - 4$ , of degree $2$ , and $A\cdot B$ has degree $6$ with leading term $-4x^{6}$ . The product rule holds without exception, because $2 \times (-2) = -4$ is not zero and no other pairing reaches degree $6$ . The sum rule can fail, and part A is exactly a case of it failing: $2 + (-2) = 0$ empties the $x^{3}$ column, so the degree drops to $2$ . The difference is the operation the two leading coefficients meet under, since addition can send two nonzero numbers to zero and multiplication cannot.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Adds coefficients within each power and leaves the powers themselves untouched. . Worth 2 points.

Names the degree from the highest power that actually survives the addition. . Worth 1 point.

Part B 4 points

Works from the two leading terms alone and says why the other products cannot reach that power. . Worth 2 points.

Multiplies the leading coefficients and adds the exponents, reporting both the degree and the whole leading term. . Worth 2 points.

Part C 5 points

Reaches a separate verdict on each of the two rules rather than one verdict covering both. . Worth 2 points. needs an explanation, not just an answer

Supports whichever rule fails with an instance from the student's own work, identifying the coefficients responsible. . Worth 2 points. needs an explanation, not just an answer

Names the operation the two leading coefficients meet under as what makes the two cases differ. . Worth 1 point.
7. A balance run forward, and a question that runs it backward . 13 points. Question 7 of 10.

An account pays $7\%$ per year compounded annually. You may use $1.07^{10} = 1.967151$ , $\log(2) = 0.3010$ and $\log(1.07) = 0.0294$ , where $\log$ is the common logarithm.
1. Part A.
  A deposit of $3200$ dollars is left in the account for $10$ years. Find the balance, to the nearest cent.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  Find, to one decimal place, the number of years in which a deposit in this account doubles, and say whether that number depends on the size of the deposit.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  Compare your answer to part A with twice the deposit, and say what your answer to part B predicts about that comparison. Then decide, with a reason, whether compounding the same $7\%$ quarterly instead of annually would make the doubling time longer or shorter.
  
  Carry your own answer forward Compare whichever balance you found in part A against twice the deposit, and read it against whichever doubling time you found in part B. The credit here is for making your own two answers agree with each other and for the reason behind the quarterly verdict.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
The answer
Part A

$6294.88$ dollars.
- $3200 \times 1.967151$ is the same value before the multiplication is carried out; $5440$ dollars is not, since that is simple interest at a flat $224$ dollars a year
Part B

About $10.2$ years, and it does not depend on the size of the deposit, because the principal cancels out of the equation.
- $\dfrac{\log 2}{\log 1.07}$ is the exact value before it is evaluated; $\log(2) - \log(1.07)$ is not the same thing, since dividing by $\log(1.07)$ is what frees the exponent
Part C

$6294.88$ dollars falls just short of $6400$ dollars, and part B explains why: doubling takes about $10.2$ years, so at $10$ years the account has not quite reached it. Quarterly compounding earns more over the same span, so it reaches double sooner and the doubling time is shorter.
Worked solution

Part A

The rate is the decimal form of the percent, and annual compounding applies the factor $1 + r$ once for each year.

$A = 3200\,(1.07)^{10} = 3200 \times 1.967151 = 6294.88$

The balance is $6294.88$ dollars, rounded to the nearest cent only at the end.

Part B

Doubling means the balance reaches $2P$ , and the principal cancels from both sides at once, which is why the answer comes out the same for every deposit.

$P(1.07)^{t} = 2P \;\Longrightarrow\; (1.07)^{t} = 2$

The unknown sits in the exponent, so take the common logarithm of both sides and let the power law bring it down to the ground.

$t\log(1.07) = \log(2) \;\Longrightarrow\; t = \frac{\log(2)}{\log(1.07)} = \frac{0.3010}{0.0294} \approx 10.2$

Part C

Twice the deposit is $2 \times 3200 = 6400$ dollars, and part A landed on $6294.88$ dollars, which is $105.12$ dollars short of it.

$6294.88 < 6400 \qquad \text{alongside} \qquad 10 < 10.2$

The two readings agree, and each is a check on the other: part B says the account needs about $10.2$ years to double, so at the ten-year mark the balance should be close to twice the deposit but not yet there, and it is.

Compounding quarterly instead pays $\tfrac{0.07}{4}$ per quarter over $4t$ quarters. Splitting the same yearly rate into more frequent steps earns a little more over any span, because interest starts earning interest sooner, so the balance climbs faster. A balance that climbs faster reaches twice its starting value earlier, so the doubling time gets shorter.

In one line

After $10$ years the balance is $3200 \times 1.07^{10} = 6294.88$ dollars. The doubling time solves $(1.07)^{t} = 2$ , giving $t = \dfrac{\log 2}{\log 1.07} \approx 10.2$ years, and the principal cancels, so that figure is the same for every deposit. The two agree: $6294.88$ dollars falls short of the $6400$ dollars that would be double, exactly because $10$ years falls short of $10.2$ . Compounding the same $7\%$ quarterly earns a little more over any span, so the balance reaches double sooner and the doubling time is shorter.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Uses the decimal rate and an exponent equal to the number of years, with the principal outside the power. . Worth 2 points.

Multiplies through and rounds only the final balance. . Worth 1 point.

States the answer as an amount of money to the nearest cent. . Worth 1 point.

Part B 5 points

Sets the balance equal to twice the principal and cancels it, rather than picking a particular deposit to work with. . Worth 1 point.

Takes a logarithm of both sides and uses the power law to bring the exponent down. . Worth 2 points.

Divides to reach a quotient of two logarithms and evaluates it to one decimal place. . Worth 1 point.

Answers the second question, saying what the cancellation of the principal means for deposits of other sizes. . Worth 1 point.

Part C 4 points

Compares the ten-year balance against twice the deposit and uses the doubling time to account for the gap, rather than reporting the two answers side by side. . Worth 2 points.

Decides the quarterly case from how the growth of the balance changes, not by guessing a direction. . Worth 2 points. needs an explanation, not just an answer
8. One tempting move, appearing in three places . 13 points. Question 8 of 10.

Three statements are in circulation:

$(1)\;\; (a+b)^{2} = a^{2} + b^{2}, \qquad (2)\;\; \log(M + N) = \log(M) + \log(N), \qquad (3)\;\; 2^{x+y} = 2^{x} + 2^{y}.$

Each one takes an addition inside a bracket and lets it pass straight through unchanged. Here $\log$ is the common logarithm, and $M$ and $N$ are positive.
1. Part A.
  Show that statement (1) is false by choosing numbers for $a$ and $b$ and evaluating both sides. Then write down the correct expansion of $(a + b)^{2}$ and name the term the false version loses.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 4 points
2. Part B.
  Do the same for statements (2) and (3): choose numbers, show that the two sides disagree, and in each case write down the true law that has that shape.
  
  Construct a counterexample Give one specific case, and show it breaks the claim. 4 points
3. Part C.
  State in one sentence what the three false statements have in common. Then explain, for each of the three true laws you have written down, why the operation on one side is not the operation on the other.
  
  Carry your own answer forward Build the summary on the counterexamples and the corrected laws you produced in parts A and B, whatever they were. The credit here is for naming the shared assumption and for saying what each true law trades, not for particular numbers.
  
  Explain why it works A sentence or two. Reasons, not steps. 5 points
The answer

Part A

With $a = 2$ and $b = 3$ : $(2+3)^{2} = 25$ while $2^{2} + 3^{2} = 13$ . The correct expansion is $(a+b)^{2} = a^{2} + 2ab + b^{2}$ , and the false version drops the cross term $2ab$ , which is $12$ here.

Part B

With $M = N = 10$ : $\log(M) + \log(N) = 2$ , which is $\log(100)$ , not $\log(20)$ , since $\log(20) = 2$ would say $10^{2} = 20$ . The true law is $\log(MN) = \log(M) + \log(N)$ . With $x = y = 3$ : $2^{6} = 64$ but $2^{3} + 2^{3} = 16$ ; the true law is $2^{x+y} = 2^{x}\cdot 2^{y}$ .

Part C

All three let an addition inside a bracket pass through unchanged. Each true law trades one operation for a different one: a square of a sum owes the cross term $2ab$ ; a logarithm turns multiplication into addition, not addition into addition; and an exponential turns addition in the exponent into multiplication.

Worked solution

Part A

Take $a = 2$ and $b = 3$ and evaluate the two sides separately.

$(2 + 3)^{2} = 5^{2} = 25, \qquad 2^{2} + 3^{2} = 4 + 9 = 13$

They are different numbers, so the statement is false. Squaring a sum means $(a+b)(a+b)$ , and every term of the first factor multiplies every term of the second, which is four products rather than two:

$(a+b)(a+b) = a^{2} + ab + ba + b^{2} = a^{2} + 2ab + b^{2}.$

The false version keeps the first and last products and discards the two cross terms, whose sum is $2ab$ . With these numbers that missing piece is $2(2)(3) = 12$ , and $13 + 12 = 25$ closes the gap exactly.

Part B

Statement (2). Take $M = N = 10$ , where both logarithms are known exactly.

$\log(10) + \log(10) = 1 + 1 = 2, \qquad \log(10 + 10) = \log(20)$

If $\log(20)$ were $2$ it would say $10^{2} = 20$ , which is false, so the two sides are different and the statement fails. What the sum $2$ actually is is $\log(100)$ , and $100 = 10 \times 10$ : the law with this shape is the product law, $\log(MN) = \log(M) + \log(N)$ , which asks for the two inputs to be MULTIPLIED. There is no law at all for the logarithm of a sum.

Statement (3). Take $x = y = 3$ .

$2^{3+3} = 2^{6} = 64, \qquad 2^{3} + 2^{3} = 8 + 8 = 16$

Again the sides disagree. The law with this shape is $2^{x+y} = 2^{x}\cdot 2^{y}$ : adding to the exponent MULTIPLIES the outputs, and $8 \times 8 = 64$ confirms it.

Part C

What they share. Each false statement assumes that an operation applied to a bracket can be handed to the terms inside it one at a time, so an addition inside comes out as an addition outside. None of the three settings works that way.

Squaring a sum. $(a+b)^{2}$ is $(a+b)(a+b)$ , and every term of the first factor multiplies every term of the second, giving four products rather than two. Two of them are the cross terms $ab$ and $ba$ , so the honest identity is $a^{2} + 2ab + b^{2}$ , and the middle term is precisely what the false version discards.

A logarithm. A logarithm is an exponent, and exponents ADD when the powers are MULTIPLIED. So the law converts a product of inputs into a sum of logarithms:

$\log(MN) = \log(M) + \log(N).$

The operation changes as it crosses the equals sign, which is the whole value of the law, and a sum of inputs is left with no rule at all.

An exponential. The same exponent fact read the other way gives $2^{x+y} = 2^{x}\cdot 2^{y}$ : adding to the input multiplies the output. Again the operation changes, addition on the left becoming multiplication on the right.

So in all three cases the true statement trades one operation for another, and a rule that hands the same operation straight through is the shape to distrust.

In one line

All three statements are false. With $a = 2$ and $b = 3$ : $(2+3)^{2} = 25$ but $2^{2} + 3^{2} = 13$ , and the true expansion $a^{2} + 2ab + b^{2}$ supplies the missing $2ab = 12$ . With $M = N = 10$ : the sum of the logarithms is $2$ , which is $\log(100)$ rather than $\log(20)$ , and the true law is $\log(MN) = \log(M) + \log(N)$ . With $x = y = 3$ : $2^{6} = 64$ but $2^{3} + 2^{3} = 16$ , and the true law is $2^{x+y} = 2^{x}\cdot 2^{y}$ . What the three share is the assumption that an addition inside a bracket passes through unchanged, when in truth each law trades one operation for a different one.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Chooses specific numbers and evaluates both sides, so that two values can actually be compared. . Worth 2 points.

Gives the correct expansion with the cross term present. . Worth 1 point.

Names the dropped term and matches its value against the gap between the two sides. . Worth 1 point.

Part B 4 points

Produces a numerical counterexample for each of the two statements, with both sides evaluated or exactly accounted for. . Worth 2 points.

States the true law behind each shape, with the correct operation on each side of it. . Worth 2 points.

Part C 5 points

Names the shared assumption as an operation being handed unchanged to the terms inside a bracket, rather than listing the three statements again. . Worth 2 points. needs an explanation, not just an answer

Says, for each true law, which operation stands on each side and why the two differ. . Worth 3 points. needs an explanation, not just an answer
9. Two functions compared over several inputs . 13 points. Question 9 of 10.

Let $f(x) = 5\cdot 2^{x}$ and $g(x) = 5 + 30x$ , both compared at the whole-number inputs $x = 0, 1, 2, 3, \ldots$ .
1. Part A.
  Evaluate $f$ and $g$ at $x = 0$ , $1$ , $2$ , $3$ , $4$ and $5$ , and state the value each rule gives at $x = 0$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  It is claimed that an exponential function with base greater than $1$ is always larger than a linear one. Decide, using your values from part A, whether that claim holds, and write down the claim about $f$ and $g$ that your values do support.
  
  Carry your own answer forward Use whichever values you produced in part A, even if they were not the expected ones, and report honestly where each function is the larger. The credit here is for testing the claim against your own numbers and for the claim your own numbers actually support.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
3. Part C.
  Look at what one step of $1$ in the input does to each rule, rather than at any table of values. Use that to explain how the two functions compare far to the right of the inputs you tried, and say why no single input can establish a claim made about every input, even though one input was enough to settle the claim in part B.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer

Part A

$f$ gives $5$ , $10$ , $20$ , $40$ , $80$ , $160$ , and $g$ gives $5$ , $35$ , $65$ , $95$ , $125$ , $155$ . Both graphs meet the vertical axis at $5$ .

Part B

It does not hold. At $x = 1$ , $f(1) = 10$ while $g(1) = 35$ , so the exponential is the smaller there, and the same holds at $x = 2$ , $3$ and $4$ . What the values support is that the exponential is EVENTUALLY larger and then stays larger, which here begins at $x = 5$ .

Part C

$g$ gains a fixed $30$ each step, while $f$ is multiplied by $2$ , so $f$ 's gain per step equals its current value and grows along with it. Once $f$ is ahead, its gain already exceeds $30$ and keeps rising, so it cannot be caught. One input reports one comparison, so it can refute a claim made about every input but can never establish one.

Worked solution

Part A

Each unit step multiplies $f$ by the base $2$ and adds $30$ to $g$ , so both rows can be built by hand.

$f: \; 5,\ 10,\ 20,\ 40,\ 80,\ 160 \qquad\qquad g: \; 5,\ 35,\ 65,\ 95,\ 125,\ 155$

The vertical intercept of each is its value at $x = 0$ . For the exponential, $f(0) = 5\cdot 2^{0} = 5\times 1 = 5$ , because any allowed base to the zero power is $1$ ; for the line, $g(0) = 5 + 30(0) = 5$ . Both begin at $5$ .

Part B

The claim fails immediately. Compare the two rows at $x = 1$ :

$f(1) = 10 \qquad \text{against} \qquad g(1) = 35$

The linear function is the larger there, and again at $x = 2$ ( $20$ against $65$ ), at $x = 3$ ( $40$ against $95$ ) and at $x = 4$ ( $80$ against $125$ ). A single input where the exponential is smaller is enough to sink a claim that says "always".

What the values do support is a claim with a timing word in it. At $x = 5$ the order changes, $160$ against $155$ , and from there the exponential is ahead. So the honest statement is that an exponential with base greater than $1$ eventually exceeds a linear function and then stays above it, not that it is larger at every input.

Part C

Look at what one unit step does to each rule, rather than at the numbers the rules produce.

$g(x+1) - g(x) = 30, \qquad f(x+1) - f(x) = 5\cdot 2^{x+1} - 5\cdot 2^{x} = 5\cdot 2^{x} = f(x)$

The linear function gains the same fixed $30$ at every step, however large it has grown. The exponential gains its own current value, so its step gain grows in proportion to the function itself. A fixed gain cannot keep pace with a growing one indefinitely, which is why the exponential must overtake.

Staying ahead follows from the same two lines. Suppose $f(x) > g(x)$ at some input with $x \ge 1$ . Then $f(x) > g(x) \ge 35 > 30$ , so at the next step $f$ gains more than $g$ does and the lead widens instead of closing, and every further step makes $f$ 's gain larger still.

A single input can never settle a claim of this kind, because it reports one comparison and nothing else. Testing $x = 1$ here would report that the linear function is larger and testing $x = 5$ would report the opposite, and neither on its own describes what happens across all inputs.

In one line

$f$ gives $5, 10, 20, 40, 80, 160$ and $g$ gives $5, 35, 65, 95, 125, 155$ , and both graphs meet the vertical axis at $5$ . The claim that an exponential with base greater than $1$ is always larger is false: at $x = 1, 2, 3$ and $4$ the linear function is the larger of the two. The corrected claim is that the exponential is eventually larger and then stays larger, which here begins at $x = 5$ . The reason is in the per-step change: $g$ gains a fixed $30$ every step while $f$ gains its own current value, so once $f$ is ahead its gain already exceeds $30$ and keeps growing.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Evaluates both rules at all six inputs, multiplying by the base for one and adding a fixed amount for the other. . Worth 2 points.

Reads each vertical intercept as the value at $x = 0$ , using that a base raised to the zero power is $1$ rather than $0$ . . Worth 1 point.

Part B 4 points

Settles the claim by exhibiting a specific input and quoting both values there, rather than asserting a verdict. . Worth 2 points. needs an explanation, not just an answer

States the claim the values support, carrying a timing word, and identifies where in the values the order changes. . Worth 2 points.

Part C 6 points

Argues from the per-step change of each rule, a fixed addition set against a multiplication, rather than from the size of the listed values. . Worth 3 points. needs an explanation, not just an answer

Accounts for the lead being permanent once it is taken, not only for the overtaking happening at all. . Worth 2 points. needs an explanation, not just an answer

Says why one input can refute a claim made about every input but can never establish one. . Worth 1 point.
10. One rate, two schedules, and a limit neither can pass . 16 points. Question 10 of 10.

A city fund holds $25000$ dollars and earns $4\%$ per year. These values are available, and not all of them are needed: $1.04^{10} = 1.480244$ , $\log(2) = 0.30103$ , $\log(1.04) = 0.01703$ , $\log(1.02) = 0.00860$ , $\log(1.01) = 0.00432$ and $\ln(2) = 0.6931$ , where $\log$ is the common logarithm and $\ln$ is the natural logarithm.
1. Part A.
  Compounded annually, find the value of the fund after $10$ years, and find the number of years in which it doubles, to one decimal place.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Compounded quarterly at the same $4\%$ , find the doubling time to one decimal place, and say how it compares with the annual figure.
  
  Carry your own answer forward Compare against whichever annual doubling time you found in part A, even if it was not the expected one. The credit here is for setting the quarterly schedule up correctly and for stating your own answer in years before you compare the two.
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
3. Part C.
  There is a shortest doubling time that no compounding schedule for this fund can beat. Say where that limit comes from, work it out, and explain what raising the number of compounding periods per year does to the doubling time relative to that limit.
  
  Carry your own answer forward Set your own two doubling times from parts A and B against the limit you work out here, and say whether they sit where your argument says they should. The credit is for locating the ceiling and for the reason it cannot be passed, not for a particular pair of earlier answers.
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
The answer
Part A

$37006.10$ dollars after $10$ years, and a doubling time of about $17.7$ years.
- $\dfrac{\log 2}{\log 1.04}$ is the doubling time before it is evaluated; $\log\!\left(\tfrac{2}{1.04}\right)$ is not the same quantity, since the division happens between the two logarithms
Part B

About $17.4$ years, roughly $0.3$ years shorter than the annual figure.

Part C

The limit is continuous compounding. As $n$ grows, $\left(1+\tfrac{r}{n}\right)^{n}$ climbs toward $e^{r}$ and no further, so no finite schedule outgrows $Pe^{rt}$ . Doubling there needs $e^{0.04t} = 2$ , giving $t = \tfrac{\ln 2}{0.04} \approx 17.3$ years, and every finite schedule stays above that.
Worked solution

Part A

Annual compounding applies the factor $1 + r = 1.04$ once for each year.

$A = 25000\,(1.04)^{10} = 25000 \times 1.480244 = 37006.10$

For the doubling time, set the balance equal to twice the principal. The principal cancels, so the answer does not depend on the $25000$ dollars at all.

$(1.04)^{t} = 2 \;\Longrightarrow\; t\log(1.04) = \log(2) \;\Longrightarrow\; t = \frac{0.30103}{0.01703} \approx 17.7$

Part B

Quarterly compounding pays $\tfrac{0.04}{4} = 0.01$ per period, and over $t$ years there are $4t$ periods. The unknown in the exponent is therefore a count of periods, not of years.

$(1.01)^{4t} = 2 \;\Longrightarrow\; 4t\log(1.01) = \log(2) \;\Longrightarrow\; 4t = \frac{0.30103}{0.00432} \approx 69.7$

Divide by $4$ to turn periods back into years, giving $t \approx 17.4$ . Set against the annual figure, that is about $0.3$ years sooner.

Part C

Where the limit comes from. Compounding $n$ times a year gives the growth factor $\left(1 + \tfrac{r}{n}\right)^{nt}$ , which regroups as $\left[\left(1 + \tfrac{r}{n}\right)^{n}\right]^{t}$ . As $n$ grows, the inner bracket rises by smaller and smaller amounts toward $e^{r}$ , because $\left(1 + \tfrac{1}{m}\right)^{m}$ crowds toward $e \approx 2.71828$ . So the balance climbs toward $Pe^{rt}$ and never passes it: the continuous case is the ceiling.

Working the limit out. Doubling under continuous compounding asks for

$Pe^{0.04t} = 2P \;\Longrightarrow\; e^{0.04t} = 2 \;\Longrightarrow\; 0.04t = \ln(2) \;\Longrightarrow\; t = \frac{0.6931}{0.04} \approx 17.3.$

The natural logarithm is the convenient one here, because $\ln$ is the logarithm to base $e$ , so $\ln\!\left(e^{0.04t}\right)$ is simply $0.04t$ .

Why the doubling time falls toward it and no further. A schedule that grows the balance faster reaches twice the principal sooner, so more frequent compounding does shorten the doubling time. But no finite schedule ever grows the balance faster than the continuous one, so none can double in less than about $17.3$ years. The three figures line up accordingly, $17.7$ for annual, $17.4$ for quarterly, and $17.3$ as the floor they are approaching, with each rise in frequency buying less than the rise before it.

In one line

Compounded annually the fund reaches $37006.10$ dollars after $10$ years and doubles in about $17.7$ years. Compounded quarterly it doubles in about $17.4$ years, roughly $0.3$ years sooner, because the rate is divided by $4$ while the exponent counts $4t$ periods. No schedule beats continuous compounding, whose one-year factor $\left(1 + \tfrac{r}{n}\right)^{n}$ climbs toward $e^{r}$ and no further: doubling there needs $e^{0.04t} = 2$ , so $t = \dfrac{\ln 2}{0.04} \approx 17.3$ years. Every finite schedule doubles in more time than that, and each rise in frequency buys less than the rise before it.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Computes the ten-year value using the decimal rate and one growth factor per year. . Worth 2 points.

Sets the balance equal to twice the principal, takes a logarithm of both sides, and reaches a quotient of two logarithms. . Worth 2 points.

Reports the value as an amount of money and the doubling time as a number of years. . Worth 1 point.

Part B 5 points

Divides the rate by $4$ and counts $4t$ periods, so the exponent solved for is a period count. . Worth 2 points.

Solves for the period count and converts it back into years before making any comparison. . Worth 2 points.

States the gap between the two doubling times, in years. . Worth 1 point.

Part C 6 points

Identifies the limiting case as continuous compounding and ties it to the one-year factor climbing toward $e^{r}$ , rather than asserting a bound. . Worth 3 points. needs an explanation, not just an answer

Solves the continuous doubling condition, using the natural logarithm because the base is $e$ . . Worth 2 points.

Places the two earlier doubling times against the limit and notes that each rise in frequency buys less than the rise before it. . Worth 1 point.

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. A pair of polynomials put through both operations 11 points

Part A 3

Combines only like powers, adding their coefficients and leaving the power itself unchanged. 2
Writes the result in descending order and names its degree from the highest power actually present. 1

Part B 4

Distributes the minus sign to every term of the second polynomial, not to the first one only. 2
Combines the like terms correctly and writes the result in descending order. 1
Names the degree from the highest power that actually survives, not from the polynomials it came from. 1

Part C 4

Locates the cause in the two leading coefficients, showing that one combination is zero and the other is not, rather than quoting a rule. 3
States the general condition with both of its halves: the same degree, and coefficients at that power that cancel. 1

2. Exponents read off, and three of them put together 12 points

Part A 3

Rewrites each logarithm as the exponential statement it stands for, rather than reaching for a formula. 2
Reports each answer as the exponent itself, including the negative value for the reciprocal input. 1

Part B 5

Factors the input into powers of the numbers whose logarithms are given, instead of looking for a rule that acts on $40$ directly. 2
Applies the product law to the two factors and the power law to the exponent, then substitutes the given values. 2
Reports a single number as the value of the logarithm. 1

Part C 4

Reaches a verdict and grounds it in what the product law actually asks of its two inputs, rather than in the line looking unfamiliar. 3
Backs the verdict with a computation, either the given values or a small case checked by hand. 1

3. Every term meeting every term, twice over 12 points

Part A 3

Forms all four products, carrying each sign into the product it belongs to. 2
Collects the two like middle terms and reports the result in descending order. 1

Part B 4

Multiplies every term of the binomial by every term of the trinomial, six products in all rather than four. 2
Combines the like powers correctly and reports the result in descending order. 2

Part C 5

States the counting rule as the product of the two term counts and matches it against both expansions. 1
Argues that the two leading terms multiply into a single top-degree term that no other pairing reaches, and that its coefficient cannot be zero. 3
Runs the argument in general letters, so it covers any two polynomials rather than only the two expanded above. 1

4. Four outputs, and the curve they belong to 13 points

Part A 4

Tests the ratios between consecutive outputs, using that the inputs are equally spaced rather than assuming it. 2
Takes the base from the constant ratio and the multiplier from the output at $x = 0$ , and writes the rule out. 2

Part B 5

Decides the direction from where the base sits relative to $1$ , not from the look of the listed numbers. 2
Evaluates the rule at $x = 0$ to get the vertical intercept. 1
States the range and the value the outputs creep toward, and keeps the two consistent with each other. 2

Part C 4

Argues from a positive base raised to a real power always being positive, rather than from the outputs merely looking small. 3
Describes what the outputs do instead, in terms of the horizontal asymptote, consistently with the sign argument. 1

5. One rate, three schedules 13 points

Part A 4

Writes the rate as a decimal and, for the quarterly schedule, divides it by $4$ while multiplying the exponent by $4$ . 2
Carries out both calculations and rounds only the final balances, not the growth factors. 1
States both answers as amounts of money to the nearest cent. 1

Part B 4

Uses $A = Pe^{rt}$ with the product of rate and time in the exponent, not the rate alone. 2
Computes the balance and then the gap against the earlier figure. 1
Reports both the balance and the gap as amounts of money. 1

Part C 5

Orders the three balances and observes that the gains between them are shrinking, not merely that they are positive. 2
Ties the ceiling to the factor for one year approaching $e^{r}$ , rather than asserting without reason that the balance is bounded. 2
Names the continuous formula as the ceiling itself, so the limit is a specific amount rather than a vague bound. 1

6. Two operations, two rules about degree 12 points

Part A 3

Adds coefficients within each power and leaves the powers themselves untouched. 2
Names the degree from the highest power that actually survives the addition. 1

Part B 4

Works from the two leading terms alone and says why the other products cannot reach that power. 2
Multiplies the leading coefficients and adds the exponents, reporting both the degree and the whole leading term. 2

Part C 5

Reaches a separate verdict on each of the two rules rather than one verdict covering both. 2
Supports whichever rule fails with an instance from the student's own work, identifying the coefficients responsible. 2
Names the operation the two leading coefficients meet under as what makes the two cases differ. 1

7. A balance run forward, and a question that runs it backward 13 points

Part A 4

Uses the decimal rate and an exponent equal to the number of years, with the principal outside the power. 2
Multiplies through and rounds only the final balance. 1
States the answer as an amount of money to the nearest cent. 1

Part B 5

Sets the balance equal to twice the principal and cancels it, rather than picking a particular deposit to work with. 1
Takes a logarithm of both sides and uses the power law to bring the exponent down. 2
Divides to reach a quotient of two logarithms and evaluates it to one decimal place. 1
Answers the second question, saying what the cancellation of the principal means for deposits of other sizes. 1

Part C 4

Compares the ten-year balance against twice the deposit and uses the doubling time to account for the gap, rather than reporting the two answers side by side. 2
Decides the quarterly case from how the growth of the balance changes, not by guessing a direction. 2

8. One tempting move, appearing in three places 13 points

Part A 4

Chooses specific numbers and evaluates both sides, so that two values can actually be compared. 2
Gives the correct expansion with the cross term present. 1
Names the dropped term and matches its value against the gap between the two sides. 1

Part B 4

Produces a numerical counterexample for each of the two statements, with both sides evaluated or exactly accounted for. 2
States the true law behind each shape, with the correct operation on each side of it. 2

Part C 5

Names the shared assumption as an operation being handed unchanged to the terms inside a bracket, rather than listing the three statements again. 2
Says, for each true law, which operation stands on each side and why the two differ. 3

9. Two functions compared over several inputs 13 points

Part A 3

Evaluates both rules at all six inputs, multiplying by the base for one and adding a fixed amount for the other. 2
Reads each vertical intercept as the value at $x = 0$ , using that a base raised to the zero power is $1$ rather than $0$ . 1

Part B 4

Settles the claim by exhibiting a specific input and quoting both values there, rather than asserting a verdict. 2
States the claim the values support, carrying a timing word, and identifies where in the values the order changes. 2

Part C 6

Argues from the per-step change of each rule, a fixed addition set against a multiplication, rather than from the size of the listed values. 3
Accounts for the lead being permanent once it is taken, not only for the overtaking happening at all. 2
Says why one input can refute a claim made about every input but can never establish one. 1

10. One rate, two schedules, and a limit neither can pass 16 points

Part A 5

Computes the ten-year value using the decimal rate and one growth factor per year. 2
Sets the balance equal to twice the principal, takes a logarithm of both sides, and reaches a quotient of two logarithms. 2
Reports the value as an amount of money and the doubling time as a number of years. 1

Part B 5

Divides the rate by $4$ and counts $4t$ periods, so the exponent solved for is a period count. 2
Solves for the period count and converts it back into years before making any comparison. 2
States the gap between the two doubling times, in years. 1

Part C 6

Identifies the limiting case as continuous compounding and ties it to the one-year factor climbing toward $e^{r}$ , rather than asserting a bound. 3
Solves the continuous doubling condition, using the natural logarithm because the base is $e$ . 2
Places the two earlier doubling times against the limit and notes that each rise in frequency buys less than the rise before it. 1

Multiple choice

Free response