Polynomials, Exponentials, and Logarithms: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

Reset the multiple-choice section?

This clears your answers and flags, and unlocks the section.

Multiple choice 0 / 20 answered

Question 1 of 20

1

What is the leading coefficient of $5x^{2} - x^{4} + 7 - 2x$ ?

Answer choices for question 1 A $-1$ B $5$ C $4$ D $7$
2

Which of these rules defines an exponential function?

Answer choices for question 2 A $f(x) = x^{3}$ B $f(x) = 3x$ C $f(x) = 3^{x}$ D $f(x) = 1^{x}$
3

Simplify $(7x^{2} - 2x + 5) - (3x^{2} + 4x - 1)$ .

Answer choices for question 3 A $4x^{2} + 2x + 4$ B $4x^{2} - 6x + 6$ C $4x^{2} - 6x + 4$ D $10x^{2} + 2x + 4$
4

Advanced. This question goes beyond core Algebra I. It is not required by the course.

Evaluate $\log_{9}(81) + \log_{5}(1)$ .

Answer choices for question 4 A $3$ B $9$ C $82$ D $2$
5

Expand $(2x - 7)(x + 3)$ .

Answer choices for question 5 A $2x^{2} - x - 21$ B $2x^{2} - 21$ C $2x^{2} + 13x - 21$ D $2x^{2} - x - 4$
6

A savings certificate pays $4\%$ compounded annually. On a deposit of $2500$ dollars, what is the balance after $2$ years? Use $1.04^{2} = 1.0816$ .

Answer choices for question 6 A $2700$ dollars B $2600$ dollars C $62500$ dollars D $2704$ dollars
7

Which statement describes the graph of $f(x) = 6\cdot\left(\tfrac{1}{3}\right)^{x}$ ?

Answer choices for question 7 A It rises from left to right, and crosses the vertical axis at $6$ . B It falls from left to right, and crosses the vertical axis at $\tfrac{1}{3}$ . C It falls from left to right, and crosses the vertical axis at $6$ . D It falls from left to right, and crosses the horizontal axis at $6$ .
8

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

Answer choices for question 8 A $x^{3} + 8x^{2} - 2x - 2$ B $x^{3} - 2x^{2} - 2$ C $x^{3} - 2x^{2} - 2x - 2$ D $x^{3} - 8x^{2} + 10$
9

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

Answer choices for question 9 A $2x^{3} + 5x^{2} - 14x - 3$ B $2x^{3} - 3x^{2} - 2x + 3$ C $2x^{3} + 5x^{2} - 10x + 3$ D $2x^{3} + 5x^{2} - 14x + 3$
10

A function has the form $f(x) = a\cdot b^{x}$ with $a > 0$ , $b > 0$ and $b \ne 1$ . Its outputs at $x = 0$ , $x = 2$ and $x = 4$ are $3$ , $12$ and $48$ . What is $b$ ?

Answer choices for question 10 A $2$ B $4$ C $16$ D $3$
11

Which expression is the exact solution of $7^{x} = 30$ ? Here $\log$ means the common logarithm, base $10$ .

Answer choices for question 11 A $x = \log(30) - \log(7)$ B $x = \dfrac{\log(7)}{\log(30)}$ C $x = \dfrac{\log(30)}{\log(7)}$ D $x = \dfrac{30}{7}$
12

What are the degree and the leading coefficient of $\left(5x^{3} + 2x^{2} - x\right) - \left(5x^{3} - 4x + 7\right)$ ?

Answer choices for question 12 A degree $3$ , leading coefficient $5$ B degree $2$ , leading coefficient $2$ C degree $6$ , leading coefficient $25$ D degree $1$ , leading coefficient $3$
13

A loan of $6000$ dollars accrues interest at $9\%$ compounded quarterly. Which expression gives the amount owed after $5$ years?

Answer choices for question 13 A $6000\,(1.0225)^{5}$ B $6000\,(1.09)^{20}$ C $6000\,(3.25)^{20}$ D $6000\,(1.0225)^{20}$
14

Simplify $(x + 5)(x - 2) - (x - 3)^{2}$ .

Answer choices for question 14 A $9x - 19$ B $-3x - 1$ C $9x - 1$ D $3x - 19$
15

Advanced. This question goes beyond core Algebra I. It is not required by the course.

The graph of $y = 4^{x}$ passes through $(2, 16)$ . Which point must lie on the graph of $y = \log_{4}(x)$ , and what is that graph's domain?

Answer choices for question 15 A $(2, 16)$ , and the domain is all real numbers. B $(16, 2)$ , and the domain is $x > 0$ . C $(16, 2)$ , and the domain is all real numbers. D $\left(\tfrac{1}{2}, \tfrac{1}{16}\right)$ , and the domain is $x > 0$ .
16

As $x$ runs further and further to the left, what value do the outputs of $f(x) = 2^{x} + 5$ creep toward, and what is the range of $f$ ?

Answer choices for question 16 A They creep toward $0$ , and the range is $y > 0$ . B They fall without bound, and the range is all real numbers. C They creep toward $5$ , and the range is $y > 5$ . D They creep toward $5$ , and the range is $y \ge 5$ .
17

A fund pays $6\%$ compounded annually. Which expression gives the number of years in which the balance reaches double the original deposit? Here $\log$ means the common logarithm.

Answer choices for question 17 A $\dfrac{\log(2)}{\log(1.06)}$ B $\dfrac{\log(2)}{\log(0.06)}$ C $\log(2) - \log(1.06)$ D $\dfrac{2}{1.06}$
18

Advanced. This question goes beyond core Algebra I. It is not required by the course.

For which real numbers $x$ is the statement $\log\!\left(x^{2}\right) = 2\log(x)$ true?

Answer choices for question 18 A Every real number $x$ . B Every real number except $x = 0$ . C No real number, since the two sides are different expressions. D Exactly the $x$ with $x > 0$ .
19

The polynomial $P$ has degree $4$ with leading coefficient $-3$ , and the polynomial $Q$ has degree $6$ with leading coefficient $\tfrac{1}{2}$ . What can be said about the degree of $P\cdot Q$ and about the degree of $P + Q$ ?

Answer choices for question 19 A $P\cdot Q$ has degree $24$ , and $P + Q$ has degree $10$ . B $P\cdot Q$ has degree $10$ , and $P + Q$ has degree $6$ . C $P\cdot Q$ has degree $10$ , and the degree of $P + Q$ cannot be settled without knowing whether the leading terms cancel. D $P\cdot Q$ has degree $10$ , and $P + Q$ has degree $10$ .
20

An exponential function $f(x) = a\cdot b^{x}$ with $a > 0$ , $b > 0$ and $b \ne 1$ satisfies $f(1) = 12$ and $f(4) = 96$ . What is $a$ ?

Answer choices for question 20 A $12$ B $\tfrac{3}{2}$ C $6$ D $2$

Submit the multiple-choice section?

Submitting locks your answers and reveals the solutions.

0/100

Where you stand

Read a polynomial's degree and its leading coefficient off the highest-degree term rather than off the first term written, putting the polynomial in standard form first. Review: Adding and Subtracting Polynomials
Add and subtract polynomials, distributing a leading minus sign to every term of the polynomial it precedes, and decide the degree of the result: at most the larger of the two degrees, and equal to it unless equal leading terms cancel. Review: Adding and Subtracting Polynomials
Multiply polynomials by forming every product of a term of one factor with a term of the other, including a binomial times a trinomial and a squared binomial that owes a cross term, and use that the degree of a product is the sum of the degrees. Review: Multiplying Polynomials
Tell an exponential function $f(x) = a\cdot b^{x}$ apart from a power or a linear function, fit one to outputs at equally spaced inputs by finding the constant ratio, and use that adding in the exponent multiplies the outputs, so an exponential eventually outgrows a linear function without being larger at every input. Review: Exponential Functions
Read the features of an exponential graph: its value at $x = 0$ , whether it rises or falls, that every output stays positive, and the value its outputs creep toward on the flat side; and read a logarithm's graph as the mirror image across $y = x$ , with domain $x > 0$ . Review: Exponential Functions , Introduction to Logarithms
Apply $A = P(1+r)^{t}$ and $A = P\left(1 + \tfrac{r}{n}\right)^{nt}$ with the rate written as a decimal and the exponent counting periods, and compare schedules against the continuous ceiling $A = Pe^{rt}$ . Review: Compound Interest
Read a logarithm as the exponent that turns the base into the input, evaluate one straight from that definition, and apply the product and power laws within their limits: there is no law for the logarithm of a sum, and $\log(M^{k}) = k\log(M)$ needs a positive $M$ . Review: Introduction to Logarithms
Solve for an unknown exponent by taking a logarithm of both sides and using the power law, reporting the answer as the quotient $\dfrac{\log y}{\log b}$ rather than a difference or a log of a quotient, and use that to find a doubling time. Review: Introduction to Logarithms , Compound Interest

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

0 of 10 completed

Progress saved in this browser.

Progress can't be saved in this browser, so your choices last for this visit only.

Problem 1 of 10

Problem 1 The replacement entry

A polynomial $R(x)$ makes $(x+2)(x^2-x+3)+R(x)=x^3+4$ for every real $x$ . Find $R$ in standard form and state its degree.
1. Hint 1
  Find the contribution of the written product before recovering the missing contribution.
2. Hint 2
  Subtract every term of that contribution from the required total.
Answer

$R(x)=-x^2-x-2$ ; degree $2$ .

Full solution

Distribution gives the six contributions $x^3$ , $-x^2$ , $3x$ , $2x^2$ , $-2x$ , and $6$ .
Thus the product is
$x^3+x^2+x+6$

Subtracting this from $x^3+4$ gives
$R(x)=-x^2-x-2$
Its highest nonzero power is $2$ .
Adding it back leaves exactly $x^3+4$ .

Answer

$R(x)=-x^2-x-2$ ; degree $2$ .

Key idea

A missing polynomial contribution is the required total minus the fully expanded known contribution.
Problem 1: Mark complete
Problem 2 The squared trinomial

Write $(x^2+3x-2)^2$ in standard form and state its degree.
1. Hint 1
  A square is the expression multiplied by itself, so every term of one copy meets every term of the other.
2. Hint 2
  Three terms times three terms give nine products; collect the ones that share a power.
Answer

$x^4+6x^3+5x^2-12x+4$ ; degree $4$ .

Full solution

The square means $(x^2+3x-2)(x^2+3x-2)$ , so each of the three terms meets each of the three terms, nine products in all.
The three products of a term with itself are
$\begin{aligned}(x^2)^2&=x^4\\(3x)^2&=9x^2\\(-2)^2&=4\end{aligned}$

Each of the three remaining pairs arises twice, once in each order, so each contributes double:
$\begin{aligned}2(x^2)(3x)&=6x^3\\2(x^2)(-2)&=-4x^2\\2(3x)(-2)&=-12x\end{aligned}$

Two contributions land on $x^2$ , and $9x^2-4x^2=5x^2$ , so the standard form is
$x^4+6x^3+5x^2-12x+4$
Each factor has degree $2$ , so the product has degree $2+2=4$ .
At $x=1$ the trinomial equals $2$ , and its square $4$ agrees with $1+6+5-12+4$ .

Answer

$x^4+6x^3+5x^2-12x+4$ ; degree $4$ .

Key idea

Squaring a multi-term expression owes every cross product as well as the squares of the individual terms.
Problem 2: Mark complete
Problem 3 The quarterly record

An account pays $8\%$ compounded quarterly, with no deposits or withdrawals after its original principal. Its balance after $9$ months is $1326.51$ dollars. Find the original principal. Round any money amount to the nearest cent.
1. Hint 1
  A quarterly schedule pays only part of the yearly rate in each period, and there are several periods each year.
2. Hint 2
  Each quarter pays $\frac{0.08}{4}$ , and $9$ months holds three of those quarters.
3. Hint 3
  Reverse the three quarterly factors to reach the principal.
Answer

Principal $1250$ dollars.

Full solution

Compounding four times a year pays
$\frac{0.08}{4}=0.02$
in each period, and $9$ months is three of those periods, so the balance is the principal multiplied by $(1.02)^3$ .

That factor is $1.061208$ , which is not zero, so the principal is
$P=\frac{1326.51}{1.061208}$
giving $1250$ dollars.
Checking forward, $1250$ dollars grows to $1275$ , then to $1300.50$ , then to $1326.51$ dollars.

Answer

Principal $1250$ dollars.

Key idea

A schedule compounded $n$ times a year divides the rate by $n$ and counts $nt$ periods in the exponent.
Problem 3: Mark complete
Problem 4 The increase records

An exponential function $f(x)=ab^x$ has $a>0$ , $b>0$ , and $b\ne1$ . Its increase from input $0$ to $1$ is $6$ , and its increase from input $1$ to $2$ is $18$ . Find its rule and sketch it on the axes in the figure. State its range and its horizontal asymptote.

Axes for the sketch: $x$ from $-3$ to $2.5$ , $y$ from $0$ to $30$ in steps of $3$ .

Text description of this figure

Blank coordinate axes ready for a sketch. The horizontal axis runs from negative three to two and a half, with a tick mark, a gridline and a number at negative three, negative two, negative one, one and two, and with the vertical axis itself standing at zero. The vertical axis runs from zero to thirty, with a tick mark, a gridline and a number every three units: three, six, nine, twelve, fifteen, eighteen, twenty one, twenty four, twenty seven and thirty, and with the horizontal axis itself standing at zero. The two scales differ, so one grid step across is one unit while one grid step up is three units. The axes are labeled x and y and the origin is labeled zero. No curve, point, asymptote or intercept mark is drawn.
1. Hint 1
  Two equal input steps have increases related by the same base factor as the outputs.
2. Hint 2
  Write the increases as $a(b-1)$ and $ab(b-1)$ , then compare them.
Answer

$f(x)=3\cdot3^x$ ; range $y>0$ ; asymptote $y=0$ ; increasing curve.

Full solution

The first increase is $a(b-1)=6$ , so it is nonzero.
The second increase is $ab(b-1)$ , so dividing it by the first cancels $a(b-1)$ and leaves the base:
$b=\frac{18}{6}$
Hence $b=3$ , and then $a(3-1)=6$ gives $a=3$ .

The rule is
$f(x)=3\cdot3^x$
It gives outputs $3$ , $9$ , and $27$ at $0$ , $1$ , and $2$ , checking both increases.

Sketch a smooth increasing curve through $(0,3)$ , $(1,9)$ , and $(2,27)$ .
It remains above the horizontal axis and approaches $y=0$ to the left, so its range is $y>0$ .

Answer

$f(x)=3\cdot3^x$ ; range $y>0$ ; asymptote $y=0$ ; increasing curve.

Key idea

For an exponential, equal input steps multiply both the outputs and their successive increases by the same factor.
Problem 4: Mark complete
Problem 5 The scale reports

Advanced. This question goes beyond core Algebra I. It is not required by the course.

A scale reports $S=\log_5(w)$ for a weight $w>0$ in grams. One item reports $3$ and a second item reports $-2$ . Find both weights, and find the total weight of the two items placed together. Then decide whether the scale's report for that total equals $3+(-2)$ , and justify your decision.
1. Hint 1
  Each report is the exponent that turns $5$ into the weight it describes.
2. Hint 2
  A negative report points to a reciprocal power of $5$ .
3. Hint 3
  Adding two reports corresponds to one operation on the weights; work out which one.
Answer

$125$ grams and $\frac{1}{25}$ gram; total $125.04$ grams; no: $3+(-2)=1$ is the report for $5$ grams, since adding reports multiplies the weights ( $125\cdot\frac{1}{25}=5$ ).

Full solution

A report is an exponent, so the first weight satisfies $5^3=w$ , giving
$w=125$
grams.

The second report is negative, and a negative exponent gives a reciprocal, so the second weight is
$5^{-2}=\frac{1}{25}$
gram, that is $0.04$ gram.

Placing the items together adds their weights:
$125+\frac{1}{25}=125.04$
grams.

The sum of the two reports is $3+(-2)=1$ , and a report of $1$ belongs to the weight $5^1=5$ grams, not to $125.04$ grams.
Adding reports corresponds to multiplying the weights, since $125\cdot\frac{1}{25}=5$ , so the report for a total weight is not the sum of the separate reports.

Answer

$125$ grams and $\frac{1}{25}$ gram; total $125.04$ grams; no: $3+(-2)=1$ is the report for $5$ grams, since adding reports multiplies the weights ( $125\cdot\frac{1}{25}=5$ ).

Key idea

Adding two logarithms multiplies their inputs, so the logarithm of a sum is not the sum of the logarithms.
Problem 5: Mark complete
Problem 6 The meeting balances

Account A starts with $500$ dollars and earns $10\%$ compounded annually. Account B starts with $750$ dollars and earns $5\%$ compounded annually. There are no other changes. In the exponential balance models for real $t\ge0$ , find the exact time, in years, when the balances are equal.
1. Hint 1
  Write one balance model for each account at the same time.
2. Hint 2
  Divide by the positive common factors to put the unknown time in one power.
Answer

$t=\frac{\log(3/2)}{\log(22/21)}$ years, equivalently $\log_{22/21}(3/2)$ years or the same quotient in natural logarithms.

Full solution

Equality of the balances gives
$500(1.10)^t=750(1.05)^t$
Dividing by the positive quantity $500(1.05)^t$ leaves one power over the other.
Equal exponents let the two powers combine, since $\frac{(1.10)^t}{(1.05)^t}=\left(\frac{1.10}{1.05}\right)^t$ , and
$\frac{1.10}{1.05}=\frac{22}{21}.$
So the equation becomes
$\left(\frac{22}{21}\right)^t=\frac32$

Taking common logarithms and applying the power law gives
$t\log(22/21)=\log(3/2)$
The logarithm of $22/21$ is positive, hence nonzero.
Dividing gives the stated positive time.

Substituting that time makes the balance ratio $1$ .
This is the meeting time in the specified real-time models; an annual credit need not occur at that exact time.

Answer

$t=\frac{\log(3/2)}{\log(22/21)}$ years, equivalently $\log_{22/21}(3/2)$ years or the same quotient in natural logarithms.

Key idea

Comparing two compounding balances reduces to one exponential equation for the ratio of their growth factors.
Problem 6: Mark complete
Problem 7 The falling readings

A quantity follows $Q(t)=ab^t$ for every real $t\ge0$ , with $a>0$ , $b>0$ and $b\ne1$ , where $t$ is time in hours. Its readings at $t=0$ , $t=1$ and $t=2$ are $250$ , $200$ and $160$ units. Find $a$ and $b$ , find the exact time at which the quantity is half its starting value, and give the first whole number of hours at which it is below half its starting value.
1. Hint 1
  Readings at equally spaced times decide whether one fixed factor acts at each step, and the reading at $t=0$ is the multiplier.
2. Hint 2
  Half the starting value leaves the unknown sitting in an exponent, where a logarithm of both sides can reach it.
3. Hint 3
  Compare the whole-hour values on either side of that exact time.
Answer

$a=250$ and $b=\frac45$ ; half at $t=\frac{\log(1/2)}{\log(4/5)}$ hours, equivalently $\frac{\log2}{\log(5/4)}$ ; first whole hour below half: $t=4$ .

Full solution

The ratios of consecutive readings are $200/250$ and $160/200$ , both equal to $0.8$ , so one fixed factor acts over each hour and
$b=\frac45$
The reading at $t=0$ is $ab^0=a$ , so $a=250$ units.

Half the starting value is $125$ units, so the exact time satisfies
$250\left(\frac45\right)^t=125$
Dividing by $250$ gives
$\left(\frac45\right)^t=\frac12.$

Taking common logarithms and applying the power law gives
$t\log\left(\frac45\right)=\log\left(\frac12\right)$
The logarithm of $4/5$ is negative, hence not zero, so dividing gives
$t=\frac{\log(1/2)}{\log(4/5)}$
which is about $3.11$ hours.

At $t=3$ the quantity is $250(0.8)^3=128$ units, still above $125$ , and at $t=4$ it is $102.4$ units, below it.
So the first whole number of hours at which it is below half is $4$ , matching an exact time between $3$ and $4$ .

Answer

$a=250$ and $b=\frac45$ ; half at $t=\frac{\log(1/2)}{\log(4/5)}$ hours, equivalently $\frac{\log2}{\log(5/4)}$ ; first whole hour below half: $t=4$ .

Key idea

A constant ratio fixes an exponential model, and the time it takes to reach a stated fraction of its starting value is then one logarithm away.
Problem 7: Mark complete
Problem 8 Tori's comparison

Nonzero polynomials $P$ and $Q$ each have degree $4$ , and $P-Q$ has degree $1$ . A nonzero polynomial $H$ has degree $3$ . Tori says $HP-HQ$ has degree $7$ since each of the two products has degree $7$ . Decide whether Tori is right and determine the actual degree.
1. Hint 1
  The leading terms of the two products need not survive subtraction.
2. Hint 2
  Factor their common polynomial $H$ and use the given degree of the difference.
Answer

Tori is wrong; the degree is $4$ .

Full solution

Each product separately has degree $3+4=7$ , but the two products share the factor $H$ , so their difference can be factored.
Distribution gives
$HP-HQ=H(P-Q)$

Both factors on the right are nonzero and have degrees $3$ and $1$ .
Thus the degree is
$3+1=4$
Tori treated a difference as though cancellation could not occur.

Answer

Tori is wrong; the degree is $4$ .

Key idea

The degree of a product adds, while the degree of a difference must account for cancellation before that rule is applied.
Problem 8: Mark complete
Problem 9 The reflected records

Advanced. This question goes beyond core Algebra I. It is not required by the course.

The figure shows an exponential curve $y=b^x$ and the line $y=x$ . Let $L$ be the function obtained by reflecting the curve across that line. Identify $L$ , then find the exact value of $L(6)+L(18)$ .

The curve $y=b^x$ and the dashed line $y=x$ .

Text description of this figure

A square coordinate grid whose horizontal and vertical axes both run from negative two to five, numbered at every whole number with equal unit lengths on the two axes. A dashed straight line runs through the origin at forty five degrees, from the lower left corner to the upper right corner, and is labeled y equals x. A smooth curve labeled y equals b to the power x rises from left to right: at the left edge it is just above the horizontal axis and comes closer to it without touching, it grows slowly at first and then steeply, and it leaves through the top of the window. Two points on that curve are drawn as filled dots and labeled with their coordinates: (0, 1) on the vertical axis, and (1, 3). No other curve, point or line is shown.
1. Hint 1
  Reflection interchanges the exponential input and output, so it gives a logarithm.
2. Hint 2
  Read the base from the curve, then combine the two positive logarithm inputs as a product.
Answer

$L(x)=\log_3(x)$ ; $L(6)+L(18)=\log_3(108)$ , equivalently $3+\log_3(4)$ .

Full solution

The marked point $(1,3)$ identifies $b=3$ .
Reflecting gives the inverse
$L(x)=\log_3(x)$
Its domain is $x>0$ , the positive range of the exponential.

The two inputs are positive, so the product law gives
$L(6)+L(18)=\log_3(108)$
Since $108=27\cdot4$ , the product law and $3^3=27$ give
$\log_3(108)=3+\log_3(4)$

The result lies between $4$ and $5$ , consistent with
$81<108<243.$

Answer

$L(x)=\log_3(x)$ ; $L(6)+L(18)=\log_3(108)$ , equivalently $3+\log_3(4)$ .

Key idea

Reflection identifies an exponential inverse, and logarithm laws combine its values through products of positive inputs.
Problem 9: Mark complete
Problem 10 The balance record

An account has a positive original principal and no deposits or withdrawals. Its balances at the ends of years $1$ , $2$ and $3$ are $385$ , $423.50$ and $465.85$ dollars. It follows either simple interest at a fixed positive annual rate or annual compounding at such a rate. Determine which model fits, find the rate and the principal, and give the limiting balance after $3$ years if that same annual rate is compounded increasingly often.
1. Hint 1
  Under simple interest the account gains the same amount every year, so weigh the two year-to-year gains against each other.
2. Hint 2
  Under annual compounding, consecutive year-end balances stand in the constant ratio $1+r$ .
3. Hint 3
  Reverse one growth factor to recover the principal, then form the continuous ceiling for the same rate and duration.
Answer

Annual compounding; rate $10\%$ ; principal $350$ dollars; limiting balance $350e^{0.3}$ dollars, about $472.45$ dollars.

Full solution

The yearly gains are $423.50-385=38.50$ dollars and $465.85-423.50=42.35$ dollars.
Simple interest pays the same amount every year, so it cannot produce two unequal gains.

Under annual compounding consecutive balances stand in the constant ratio $1+r$ , and here
$\frac{423.50}{385}=1.10$
The later pair agrees, since $465.85/423.50=1.10$ as well, so $r=0.10$ , or $10\%$ .

Reversing one year of growth gives the principal
$P=\frac{385}{1.10}$
which is $350$ dollars.
Checking forward, $350$ dollars grows to $385$ , then to $423.50$ , then to $465.85$ dollars.

For this principal, rate and three-year duration, increasing the number of periods without bound approaches
$A=350e^{0.3}$
dollars, about $472.45$ dollars, which stands just above the annual balance as a ceiling must.

Answer

Annual compounding; rate $10\%$ ; principal $350$ dollars; limiting balance $350e^{0.3}$ dollars, about $472.45$ dollars.

Key idea

Equal yearly gains mark simple interest and a constant yearly ratio marks compounding, and no finite schedule passes the continuous ceiling.
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Algebra I. You can skip it. → Continue to Chapter 14: Special Functions →

← Previous Chapter Review Next → Star problems Advanced. This problem set goes beyond core Algebra I. You can skip it.