Chapter test · nothing is marked until you submit

Exponential and Logarithmic Functions: 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, 118 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 table with one entry missing, and the base it pins down . 12 points. Question 1 of 10.

An exponential function $f(x) = a\cdot b^{x}$ is sampled at four consecutive integer inputs. Three readings survive: $f(1)=20$ , $f(2)=100$ , and $f(4)=2500$ ; the fourth, $f(3)$ , was lost.
1. Part A.
  First use the ratio test on the two readings one step apart, $f(1)$ and $f(2)$ , to find $b$ directly. Then check that this same value of $b$ is consistent with the wider three-step gap between $f(1)$ and $f(4)$ , and use $f(1)$ to find $a$ and write the rule for $f(x)$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Using your rule, find the missing table value $f(3)$ , and verify it is consistent with both neighboring entries by checking that the one-step ratio test holds across each gap.
  
  Carry your own answer forward Use your own rule for $f(x)$ from part A to fill in the missing entry.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  A classmate claims that since $f(1)=20$ is positive, $b$ could just as easily have been $-5$ instead of $5$ , because negative signs might cancel somewhere in the division. Explain precisely where the algebra in part A rules out a negative value of $b$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$b=5$ from $f(2)/f(1)$ ; the wider gap check gives $b^{3}=125$ , consistent; $a=4$ , so $f(x)=4\cdot5^{x}$ .

Part B

$f(3)=500$ ; $f(3)/f(2)=5=b$ and $f(4)/f(3)=5=b$ , both consistent.

Part C

The division in part A produces $b^{3}=125$ , a positive number. Checking directly, $(-5)^{3}=-125\neq125$ , so $b=-5$ fails the equation outright, and separately, an exponential base is required to be positive.

Worked solution

Part A

The ratio test on two readings one step apart gives $b$ directly, since the exponent gap is exactly $1$ .

$\frac{f(2)}{f(1)} = b^{1} = \frac{100}{20} = 5 \quad\Longrightarrow\quad b=5$

Check this against the wider three-step gap between $f(1)$ and $f(4)$ : $b^{3}$ should equal $\dfrac{f(4)}{f(1)}$ .

$b^{3} = 5^{3} = 125 = \frac{2500}{20} = \frac{f(4)}{f(1)}$

The two readings agree, confirming a single exponential fits all three data points. Substitute back into $f(1)=20$ : $5a=20$ , so $a=4$ . Thus $f(x)=4\cdot5^{x}$ .

Part B

Substitute $x=3$ into the rule found in part A.

$f(3) = 4\cdot5^{3} = 4\cdot125 = 500$

Check consistency: $\dfrac{f(3)}{f(2)}=\dfrac{500}{100}=5=b$ , and $\dfrac{f(4)}{f(3)}=\dfrac{2500}{500}=5=b$ . Both one-step ratios equal $b$ , confirming the missing value fits the same exponential as every other entry in the table.

Part C

The step that pins down $b$ is $b^{3}=\dfrac{f(4)}{f(1)}=125$ . Checking $b=-5$ directly: $(-5)^{3}=-125$ , which is not equal to $125$ , so $b=-5$ does not even satisfy this equation; the classmate's algebra does not lead there at all.

Separately, even if some other equation had produced $b^{3}=-125$ , the base of an exponential function is required to be positive, so only the positive real cube root is ever accepted as $b$ .

$b^{3}=125 \quad\Longrightarrow\quad b=5 \text{ (the positive cube root)}, \qquad (-5)^{3}=-125\neq125$

In one line

$b=5$ (from the one-step ratio $f(2)/f(1)$ , confirmed by the three-step check $b^{3}=125$ ) and $a=4$ , so $f(x)=4\cdot5^{x}$ ; the missing table value is $f(3)=500$ , consistent with both neighboring ratios; and $b=-5$ is ruled out because $(-5)^{3}=-125\neq125$ fails the equation $b^{3}=125$ outright, and even apart from that a base must be positive.

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 ratio test on the two readings one step apart, $f(2)/f(1)$ , to find $b$ directly. . Worth 2 points.

Checks the resulting $b$ against the wider three-step gap between $f(1)$ and $f(4)$ , and finds $a$ from $f(1)$ . . Worth 2 points.

Part B 4 points

Substitutes $x=3$ into the rule found in part A to compute the missing value. . Worth 2 points.

Verifies the missing value against both neighboring one-step ratios, confirming the table is consistent throughout. . Worth 2 points.

Part C 4 points

Checks $b=-5$ directly against the equation $b^{3}=125$ and finds $(-5)^{3}=-125\neq125$ , so it fails outright. . Worth 2 points. needs an explanation, not just an answer

States that even apart from that failure, an exponential function's base must be positive, which rules out $b=-5$ on its own. . Worth 2 points.
2. A table that pins down a base no real exponential can have . 12 points. Question 2 of 10.

A student proposes to fit an exponential model $k(x)=a\cdot c^{x}$ to the table $k(0)=5$ , $k(1)=-15$ , $k(2)=45$ , $k(3)=-135$ .
1. Part A.
  Use the ratio test to show a single value of $c$ fits every one-step gap in the table, and find that value. Then, treating $c$ as a proposed exponential base, evaluate $c^{1/2}$ and explain why no real number could ever serve as its value.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
2. Part B.
  Show that for ANY negative number $d$ , $d^{1/2}$ cannot be a real number, using the same squaring argument as in part A, and state what this means for the set of allowed exponential bases.
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 4 points
3. Part C.
  A classmate proposes the base $c=0$ instead, arguing that since $0^{x}=0$ for every POSITIVE $x$ , this base avoids the kind of failure found in part A entirely. Identify the specific real input where $c=0$ still breaks down, and explain why that failure is a fundamentally different KIND of failure from the one in part A.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

$c=-3$ fits every gap; $c^{1/2}$ would need to square to $-3$ , and no real number squares to a negative, so $c^{1/2}$ does not exist.

Part B

For any $d<0$ , $\left[d^{1/2}\right]^{2}$ would have to equal $d$ , a negative number, but a real square is never negative, so $d^{1/2}$ is never real; this means the set of allowed exponential bases can never include a negative number.

Part C

$0^{x}$ breaks down at $x=-1$ (and every negative $x$ ), since it would require dividing by $0^{1}=0$ ; that is a division-by-zero failure, not a missing-real-root failure like the one in part A.

Worked solution

Part A

Compute the three one-step ratios in the table.

$\frac{-15}{5}=-3, \qquad \frac{45}{-15}=-3, \qquad \frac{-135}{45}=-3$

All three agree, so a single value $c=-3$ fits every entry in the table.

The law of exponents forces $c^{1}=\left[c^{1/2}\right]^{2}$ , since $\tfrac12+\tfrac12=1$ . So $\left[c^{1/2}\right]^{2}$ would have to equal $c=-3$ . A square of a real number is never negative, so no real number can play the role of $c^{1/2}$ .

$c^{1}=\left[c^{1/2}\right]^{2}=-3 \quad\Longrightarrow\quad \text{impossible for real } c^{1/2}$

Part B

Let $d<0$ be any negative number, and suppose $d^{1/2}$ were a real number. By the same law of exponents used in part A,

$\left[d^{1/2}\right]^{2} = d^{1} = d$

But the left side is a real number squared, which is never negative, while the right side $d$ is negative by assumption. That contradiction holds for every negative $d$ , not just $c=-3$ , so $d^{1/2}$ is never a real number when $d$ is negative. Since an exponential function must have SOME real value at every real input, including $x=\tfrac12$ , no negative number can ever serve as its base.

Part C

The classmate is right that $0^{x}=0$ works fine for positive $x$ , so this base avoids exactly the failure found in part A: there is no missing real root anywhere, since $0$ raised to any positive real power is perfectly well-defined.

But $c=0$ breaks down elsewhere. Consider $x=-1$ : by the law of exponents, $0^{-1}$ would have to equal $\dfrac{1}{0^{1}}=\dfrac10$ , division by zero, which is undefined. The same failure repeats at every negative $x$ .

This is a fundamentally different KIND of failure from part A's. There, the obstacle was that squaring a real number can never produce a negative result, a genuine gap in what real numbers can do. Here, the obstacle is dividing by zero, an operation that is simply never allowed, regardless of what number sits on top. Both failures disqualify their base from ever serving as an exponential base, but for two structurally different reasons.

$0^{-1} = \frac{1}{0^{1}} = \frac{1}{0} \ \text{(undefined)} \quad\text{vs.}\quad c^{1/2} \text{ squares to a negative (impossible)}$

In one line

The ratio test shows $c=-3$ fits every entry in the table, but $c^{1/2}$ would have to square to $-3$ , which is impossible, and the same argument rules out EVERY negative base, not just $-3$ . A base of $c=0$ fails for a different reason: it is defined for positive exponents but breaks down at every negative exponent, since it would require dividing by zero, a fundamentally different kind of failure from the missing-real-root problem.

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 ratio test to confirm a single value of $c$ fits every gap in the table. . Worth 1 point.

Uses the law of exponents to write $c^{1}$ as the square of $c^{1/2}$ . . Worth 2 points.

Concludes correctly that no real number squares to a negative number, so $c^{1/2}$ does not exist. . Worth 1 point. needs an explanation, not just an answer

Part B 4 points

Sets up the general squaring identity $\left[d^{1/2}\right]^{2}=d$ for an arbitrary $d<0$ . . Worth 2 points.

Concludes the contradiction holds for EVERY negative $d$ , not just $c=-3$ , and connects it to the base restriction on an exponential function. . Worth 2 points. needs an explanation, not just an answer

Part C 4 points

Identifies a specific real input (any negative $x$ , such as $x=-1$ ) where $0^{x}$ breaks down. . Worth 2 points.

Explains that this is a division-by-zero failure, a fundamentally different kind of breakdown from the missing-real-root failure in part A. . Worth 2 points. needs an explanation, not just an answer
3. Reading one graph's ceiling off the other's floor . 11 points. Question 3 of 10.

The graph of $y = 8^{x}$ passes through the points $\left(-\tfrac13,\ \tfrac12\right)$ and $(2,\ 64)$ .
1. Part A.
  State the two points that must lie on the graph of $y=\log_8 x$ , using the reflection across $y=x$ alone.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points
2. Part B.
  Take the point you found in part A whose SECOND coordinate is a whole number, and verify it directly from the DEFINITION of a logarithm (not the reflection). Then use the definition to evaluate $\log_8 \tfrac18$ .
  
  Carry your own answer forward Use your own point from part A for the verification, then evaluate the new logarithm from the definition directly.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Explain why the domain of $y=\log_8 x$ is exactly the RANGE of $y=8^{x}$ , rather than some other set, using only the fact that reflecting across $y=x$ swaps the roles of input and output.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$\left(\tfrac12,-\tfrac13\right)$ and $(64,2)$ .

Part B

$8^{2}=64$ confirms $(64,2)$ ; $\log_8\tfrac18=-1$ .

Part C

Reflection swaps every point's coordinates, so whatever set of values served as OUTPUTS (the range) for $y=8^x$ becomes the set of INPUTS (the domain) for $y=\log_8x$ , and that swap is exactly what defines an inverse function's domain and range.

Worked solution

Part A

Reflecting across $y=x$ swaps each point's coordinates.

$\left(-\frac13,\ \frac12\right)\to\left(\frac12,\ -\frac13\right), \qquad (2,\ 64)\to(64,\ 2)$

Part B

By the definition, $(64,2)$ lies on $y=\log_8x$ exactly when $8^{2}=64$ , which is true, confirming the point independently of the reflection argument.

For the new value, ask the defining question: eight to the what gives one eighth? A number smaller than $1$ needs a negative exponent.

$8^{-1}=\frac18 \quad\Longrightarrow\quad \log_8\frac18=-1$

Part C

A point $(p,q)$ on $y=8^{x}$ reflects to $(q,p)$ on $y=\log_8x$ . The first coordinate of a point is its input, and the second is its output.

Before reflection, $p$ ranges over every possible input of $8^{x}$ (its domain) and $q$ ranges over every possible output (its range). After reflection, $q$ becomes the first coordinate, that is, the input, of the new graph, and $p$ becomes its output.

$\text{domain of } \log_8x = \{q : q \text{ was an output of } 8^{x}\} = \text{range of } 8^{x}$

So the swap of coordinates under reflection is precisely why an inverse function's domain is its original function's range, with no separate argument needed.

In one line

$\left(\tfrac12,-\tfrac13\right)$ and $(64,2)$ lie on $y=\log_8x$ ; the definition confirms $8^{2}=64$ , and $\log_8\tfrac18=-1$ since $8^{-1}=\tfrac18$ ; the domain of $\log_8x$ equals the range of $8^{x}$ because reflection across $y=x$ swaps every point's input and output roles.

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

Swaps the coordinates of both given points correctly. . Worth 2 points.

Reports the two resulting points clearly, without recomputing either one from the definition of a logarithm. . Worth 1 point.

Part B 4 points

Confirms $8^{2}=64$ as an independent check of the point from part A. . Worth 1 point.

Evaluates $\log_8\tfrac18$ correctly using the definition. . Worth 2 points.

Reports both results as clean exact values. . Worth 1 point.

Part C 4 points

Explains that reflection swaps the input and output roles of every point's coordinates. . Worth 2 points. needs an explanation, not just an answer

Connects that swap explicitly to the general fact that an inverse function's domain equals the original's range, and vice versa. . Worth 2 points.
4. Two expressions that look alike but need different checks . 12 points. Question 4 of 10.

This question checks the two cancellation laws side by side, then asks about the domain condition each one carries.
1. Part A.
  Simplify $\log_9\!\left(9^{-14}\right)$ and $11^{\log_{11} 53}$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  State, without computing anything, whether $\log_{15}\!\left(15^{-2/3}\right)$ and $6^{\log_6(-53)}$ are each defined, and give the value of whichever one is.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Let $j(x)=\log_5\!\left(5^{\,x^{2}-9}\right)$ and $k(x)=5^{\log_5(x^{2}-9)}$ . Using the two cancellation laws, find the domain of each composite function, and explain why the two domains come out different even though $j$ and $k$ look like mirror images of each other.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$\log_9\!\left(9^{-14}\right)=-14$ and $11^{\log_{11}53}=53$ .

Part B

$\log_{15}\!\left(15^{-2/3}\right)$ is defined and equals $-\tfrac23$ ; $6^{\log_6(-53)}$ is not defined, since $\log_6(-53)$ does not exist.

Part C

Domain of $j$ is all reals, since $\log_5(5^{t})=t$ needs no restriction on $t=x^{2}-9$ . Domain of $k$ is $x<-3$ or $x>3$ , since $5^{\log_5u}=u$ needs $u=x^{2}-9>0$ first.

Worked solution

Part A

The first matches $\log_b(b^{t})=t$ , true for every real $t$ .

$\log_9\!\left(9^{-14}\right)=-14$

The second matches $b^{\log_bx}=x$ , true for every $x>0$ .

$11^{\log_{11}53}=53$

Part B

The first is an instance of $\log_b(b^{t})=t$ , which holds for EVERY real $t$ , including the fraction $-\tfrac23$ , so it is defined and equals $-\tfrac23$ .

The second needs $\log_6(-53)$ to exist first, and a logarithm's argument must be positive; $-53$ is not, so $\log_6(-53)$ does not exist and the whole expression $6^{\log_6(-53)}$ is undefined.

$\log_{15}\!\left(15^{-2/3}\right) = -\frac23, \qquad 6^{\log_6(-53)} \text{ undefined}$

Part C

For $j(x)=\log_5\!\left(5^{\,x^{2}-9}\right)$ , the identity $\log_5(5^{t})=t$ holds for EVERY real $t$ , with no restriction, since $5^{t}$ is automatically a positive, legal input to the logarithm. Here $t=x^{2}-9$ can be any real number as $x$ ranges over all reals, so $j(x)=x^{2}-9$ for every real $x$ : the domain of $j$ is all real numbers.

For $k(x)=5^{\log_5(x^{2}-9)}$ , the identity $5^{\log_5u}=u$ needs its INNER piece $\log_5(x^{2}-9)$ to exist first, which requires $x^{2}-9>0$ .

$x^{2}-9>0 \quad\Longrightarrow\quad (x-3)(x+3)>0 \quad\Longrightarrow\quad x<-3 \text{ or } x>3$

So the domain of $k$ is $x<-3$ or $x>3$ , strictly smaller than all reals.

Although $j$ and $k$ look like mirror images built from the same two expressions, their domains differ because one cancellation law's inner piece ( $5^{t}$ ) is automatically positive for every real $t$ , while the other's inner piece ( $\log_5(x^{2}-9)$ ) exists only when $x^{2}-9$ is already positive. That asymmetry, not the surface resemblance of the two formulas, is what decides the domain.

In one line

$\log_9\!\left(9^{-14}\right)=-14$ and $11^{\log_{11}53}=53$ ; $\log_{15}\!\left(15^{-2/3}\right)=-\tfrac23$ is defined, but $6^{\log_6(-53)}$ is not, since $\log_6(-53)$ fails; applying the same two laws to the composite functions $j(x)=\log_5(5^{x^{2}-9})$ and $k(x)=5^{\log_5(x^{2}-9)}$ shows their domains differ even though the formulas look like mirror images: $j$ 's domain is all reals, since $5^{t}$ is always a valid positive input, while $k$ 's domain is $x<-3$ or $x>3$ , since $\log_5(x^{2}-9)$ needs $x^{2}-9>0$ first.

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

Identifies which cancellation law applies to each expression before simplifying. . Worth 2 points.

Evaluates both expressions correctly. . Worth 1 point.

Reports both results as plain numbers, with no leftover exponent notation. . Worth 1 point.

Part B 4 points

Correctly classifies each expression as defined or not. . Worth 2 points.

Ties the undefined case specifically to the negative argument of the inner logarithm. . Worth 2 points. needs an explanation, not just an answer

Part C 4 points

Correctly finds the domain of $j$ is all real numbers and the domain of $k$ is $x<-3$ or $x>3$ . . Worth 2 points.

Explains that the difference in domains traces to which cancellation law's inner piece is automatically positive. . Worth 2 points. needs an explanation, not just an answer
5. Counting the digits of a number too large to write down . 13 points. Question 5 of 10.

A computer scientist wants to know how many decimal digits the number $6^{50}$ has, without multiplying it out.
1. Part A.
  Use the power rule to write $\log_{10}\!\left(6^{50}\right)$ in terms of $\log_{10}6$ , then evaluate it using $\log_{10}6\approx0.77815$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Using your value from part A, state how many decimal digits $6^{50}$ has, and explain the rule connecting a whole number's digit count to where its logarithm falls.
  
  Carry your own answer forward Use your own value of $\log_{10}\!\left(6^{50}\right)$ from part A to determine the digit count.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
3. Part C.
  A classmate argues that whenever $50\log_{10}6$ 's decimal part happens to be closer to $1$ than to $0$ , the digit count should be found by ROUNDING to the nearest integer rather than by the floor-plus-one rule from part B. Explain why this reasoning is wrong in general, even though it happens to agree with the correct answer for THIS particular calculation, by describing (in general terms, no new numbers needed) a case where rounding to the nearest integer would give the WRONG digit count.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 4 points
The answer

Part A

$\log_{10}\!\left(6^{50}\right)=50\log_{10}6\approx38.9075$ .

Part B

$6^{50}$ has $39$ digits, since a positive integer $N$ has $d$ digits exactly when $d-1\le\log_{10}N<d$ , and $38\le38.9075<39$ .

Part C

Rounding to the nearest integer answers a different question than the digit-count rule; for any logarithm with a fractional part below $0.5$ , rounding to the nearest integer would round DOWN, undercounting the digits by one.

Worked solution

Part A

The power rule brings the exponent to the front.

$\log_{10}\!\left(6^{50}\right)=50\log_{10}6 \approx 50(0.77815)=38.9075$

Part B

A positive integer $N$ has $d$ digits exactly when $10^{d-1}\le N<10^{d}$ , which is the same as $d-1\le\log_{10}N<d$ .

Here $\log_{10}\!\left(6^{50}\right)\approx38.9075$ , which sits between $38$ and $39$ , so $d-1=38$ and $d=39$ .

$38 \le 38.9075 < 39 \quad\Longrightarrow\quad 6^{50} \text{ has } 39 \text{ digits}$

Part C

The digit-count rule is $d-1\le\log_{10}N<d$ : the digit count is always the FLOOR of the logarithm plus one, regardless of how close the decimal part is to $0$ or to $1$ .

Rounding to the nearest integer answers a different question, which of the two nearby integers the decimal is numerically closer to. The two rules agree only when the decimal part happens to exceed $0.5$ , as it did here with $0.9075$ .

Consider any logarithm of the general form $d-1+0.3$ , whose decimal part is below $0.5$ . The digit-count rule still gives $d$ digits, but rounding to the nearest integer would round DOWN to $d-1$ , undercounting by exactly one.

$d-1 \le d-1+0.3 < d \quad\text{(true digit count } d\text{)}, \qquad \text{nearest integer} = d-1 \ne d$

In one line

$\log_{10}\!\left(6^{50}\right)=50\log_{10}6\approx38.9075$ , so $6^{50}$ has $39$ digits, since $38\le38.9075<39$ ; rounding to the nearest integer happens to agree here only because the decimal part exceeds $0.5$ , but for any logarithm with decimal part below $0.5$ rounding would undercount the digits by 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 5 points

Applies the power rule to bring the exponent to the front. . Worth 2 points.

Computes the numeric value correctly. . Worth 2 points.

Reports the value as a decimal approximation consistent with the given approximation of $\log_{10}6$ . . Worth 1 point.

Part B 4 points

States the digit-count rule connecting a logarithm's trapped value to a digit count. . Worth 1 point.

Correctly reads off the digit count from the trapped logarithm. . Worth 2 points.

Shows the distinction between $d-1$ and $d$ explicitly, rather than only stating the final digit count. . Worth 1 point.

Part C 4 points

Explains that rounding answers a different question than flooring the logarithm and adding one. . Worth 2 points. needs an explanation, not just an answer

Gives a general case, a logarithm with a decimal part below $0.5$ , where rounding would undercount the digits. . Worth 2 points.
6. Finding the one broken step in someone else's solution . 12 points. Question 6 of 10.

A student's full attempt to solve $\log_2(x+9) + \log_2(x-3) = 6$ is shown below. Every step uses a valid METHOD for this kind of equation, except exactly one, which contains a planted algebra error.

Step 1 (domain): needs $x+9>0$ and $x-3>0$ , so $x>-9$ and $x>3$ , giving $x>3$ . Step 2 (combine): $\log_2\big((x+9)(x-3)\big)=6 \Rightarrow (x+9)(x-3)=2^{6}=64$ . Step 3 (expand): $(x+9)(x-3)=x^{2}-27$ , so $x^{2}-27=64 \Rightarrow x^{2}=91 \Rightarrow x=\sqrt{91}$ . Step 4: Since $\sqrt{91}\approx9.54>3$ , the domain is satisfied, so the student reports $x=\sqrt{91}$ as the final answer.
1. Part A.
  Find the exact planted error in the student's work (name the step and describe the mistake precisely), and correctly expand $(x+9)(x-3)$ .
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 4 points
2. Part B.
  Using the corrected expansion, solve the equation for every candidate value of $x$ , and test each against the domain from Step 1 to determine which is genuine.
  
  Carry your own answer forward Use your own corrected expansion from part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Two different checks appear in this problem: the domain check in Step 1 that would reject the extraneous candidate you found in part B, and a full substitution check (plugging a candidate back into the ORIGINAL equation) that was never performed. Explain why only the SECOND check, not the domain check, would have revealed the Step 3 error, given that $\sqrt{91}$ genuinely satisfies the domain $x>3$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

The error is in Step 3: the student dropped the two cross terms when expanding, writing $(x+9)(x-3)=x^{2}-27$ instead of the correct $x^{2}+6x-27$ .

Part B

$x=7$ or $x=-13$ ; only $x=7$ satisfies the domain $x>3$ , so $x=-13$ is extraneous.

Part C

A domain check only tests whether $x=\sqrt{91}$ makes both arguments positive, which it does, so it reveals nothing. A full substitution recomputes $\log_2(x+9)+\log_2(x-3)$ at $x=\sqrt{91}$ and finds it does not equal $6$ , exposing the arithmetic error upstream of the domain.

Worked solution

Part A

The domain in Step 1 and the combination in Step 2 are both correct. The error is in Step 3's expansion. FOIL requires all four products:

$(x+9)(x-3) = x^{2} - 3x + 9x - 27 = x^{2} + 6x - 27$

The student kept the first and last terms but dropped BOTH cross terms ( $-3x$ and $9x$ ), which cancel only if they are equal and opposite, and they are not here ( $-3x+9x=6x\neq0$ ).

Part B

$x^{2}+6x-27=64 \;\Longrightarrow\; x^{2}+6x-91=0 \;\Longrightarrow\; (x-7)(x+13)=0 \;\Longrightarrow\; x=7 \text{ or } x=-13$

Testing against the domain $x>3$ from Step 1: $x=7>3$ , genuine. $x=-13$ fails the domain outright, since $-13$ is nowhere near $x>3$ ; extraneous.

Part C

The domain check in Step 1 only asks one question: are the two arguments positive at this candidate? For $x=\sqrt{91}\approx9.54$ , both $x+9$ and $x-3$ are positive, so this check passes cleanly and gives no indication that anything upstream went wrong.

A full substitution check asks a different, stronger question: does the candidate actually satisfy the ORIGINAL equation, arithmetic and all? Plugging $x=\sqrt{91}$ back in would require recomputing $\log_2(x+9)+\log_2(x-3)$ from scratch and comparing it to $6$ ; since $\sqrt{91}$ came from the wrong expansion in Step 3, it is not a genuine root, and this recomputation would fail to reach $6$ , exposing the error immediately.

$\text{domain check: is } x+9>0 \text{ and } x-3>0? \quad\text{vs.}\quad \text{substitution check: does } \log_2(x+9)+\log_2(x-3)=6?$

The domain check catches candidates that are OUTSIDE the equation's domain, like $x=-13$ ; it cannot catch a candidate that is inside the domain but simply wrong because of an earlier arithmetic slip. Only a full substitution check catches that kind of error.

In one line

The planted error is in Step 3: $(x+9)(x-3)$ should expand to $x^{2}+6x-27$ , not $x^{2}-27$ ; using the corrected expansion, $x^{2}+6x-91=0$ gives $x=7$ or $x=-13$ , and only $x=7$ satisfies the domain $x>3$ . The domain check alone would not have caught the Step 3 error, since $\sqrt{91}$ genuinely satisfies $x>3$ ; only a full substitution back into the original equation, which $\sqrt{91}$ fails, would expose an algebra mistake like this 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

Correctly identifies Step 3 as the location of the error. . Worth 1 point.

Correctly re-expands $(x+9)(x-3)$ using all four FOIL products. . Worth 2 points.

Names the specific mistake (dropping the two cross terms) rather than only stating that Step 3 is wrong. . Worth 1 point.

Part B 4 points

Solves the corrected quadratic and reports both candidate values of $x$ . . Worth 2 points.

Tests both candidates against the domain from Step 1 and correctly classifies each. . Worth 2 points.

Part C 4 points

Explains that the domain check only tests positivity and passes cleanly for $x=\sqrt{91}$ , revealing nothing. . Worth 2 points. needs an explanation, not just an answer

Explains that a full substitution recomputes the original equation and would fail at $x=\sqrt{91}$ , catching the arithmetic error. . Worth 2 points.
7. A quadratic hiding inside a repeated logarithm . 13 points. Question 7 of 10.

Let $u=\log_6 x$ . This turns $(\log_6 x)^{2} - 7\log_6 x + 10 = 0$ into an ordinary quadratic in $u$ .
1. Part A.
  Rewrite the equation as a quadratic in $u$ and solve it for every value of $u$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 5 points
2. Part B.
  Undo the substitution for each value of $u$ found in part A, solving exactly for $x$ , and verify both in the original equation.
  
  Carry your own answer forward Use whichever two values of $u$ you found in part A, even if they differ from the ones intended.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  For a DIFFERENT equation, $(\log_6 x)^{2} - 7\log_6 x + 10 = 0$ , consider rewriting $(\log_6 x)^{2}$ as $\log_6(x^{2})$ , which would turn the equation into $\log_6(x^{2}) - 7\log_6 x + 10=0$ . Determine whether this rewrite is valid, and explain, using a specific numeric example, why $(\log_6 x)^{2}$ and $\log_6(x^{2})$ are not the same expression.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 4 points
The answer

Part A

$u^{2}-7u+10=0$ , giving $u=2$ or $u=5$ .

Part B

$u=2$ gives $x=36$ ; $u=5$ gives $x=7776$ . Both check.

Part C

$(\log_6x)^{2}$ squares the VALUE of the logarithm, while $\log_6(x^{2})$ takes the logarithm of $x^{2}$ ; e.g. at $x=6$ , $(\log_66)^{2}=1^{2}=1$ but $\log_6(6^{2})=2$ , so the two expressions are not equal.

Worked solution

Part A

Substituting $u=\log_6 x$ turns every term into a term in $u$ .

$u^{2}-7u+10=0 \;\Longrightarrow\; (u-2)(u-5)=0 \;\Longrightarrow\; u=2 \text{ or } u=5$

Part B

Convert each value of $u$ back using $x=6^{u}$ .

$u=2: \; x=6^{2}=36, \qquad u=5: \; x=6^{5}=7776$

Check: at $x=36$ , $\log_6 36=2$ , so $4-14+10=0$ . At $x=7776$ , $\log_6 7776=5$ , so $25-35+10=0$ . Both survive.

Part C

$(\log_6 x)^{2}$ means: first evaluate $\log_6 x$ , then square the resulting NUMBER. $\log_6(x^{2})$ means something entirely different: first square $x$ , then take the logarithm of that new, larger number. These are two different orders of operation applied to two different things, and nothing in the power rule justifies swapping them; the power rule moves an exponent OUTSIDE a logarithm ( $\log_b(x^{n}) = n\log_b x$ ), it never lets you move an exponent from outside the logarithm to inside its argument.

A single numeric check settles it. Take $x=6$ :

$(\log_6 6)^{2} = 1^{2} = 1, \qquad \log_6\!\left(6^{2}\right) = \log_6 36 = 2$

Since $1 \ne 2$ , the two expressions genuinely disagree at this input, confirming the proposed rewrite is false, not merely differently written.

In one line

$(\log_6x)^{2}-7\log_6x+10=0$ becomes $u^{2}-7u+10=0$ under $u=\log_6x$ , giving $u=2$ or $u=5$ , so $x=36$ or $x=7776$ , both checking. The proposed rewrite $(\log_6x)^{2}=\log_6(x^{2})$ is false: the left side squares the VALUE of the logarithm while the right side takes the logarithm of $x^{2}$ ; at $x=6$ these give $1$ and $2$ respectively, confirming they disagree.

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

Substitutes correctly, recognizing $(\log_6x)^{2}$ as $u^{2}$ . . Worth 2 points.

Factors the quadratic and reports both values of $u$ . . Worth 2 points.

States both values of $u$ clearly as intermediate results, distinguishing them from the values of $x$ still to be found. . Worth 1 point.

Part B 4 points

Correctly undoes the logarithmic substitution for both values of $u$ . . Worth 2 points.

Verifies both resulting values of $x$ in the original equation. . Worth 2 points.

Part C 4 points

Explains the conceptual difference: $(\log_6x)^{2}$ squares the value of the logarithm, while $\log_6(x^{2})$ takes the logarithm of $x^{2}$ . . Worth 2 points. needs an explanation, not just an answer

Produces a numeric counterexample (e.g. $x=6$ ) showing the two expressions disagree. . Worth 2 points.
8. A colony that triples on its own schedule . 10 points. Question 8 of 10.

A bacteria colony starts at $40$ cells and triples in size every $5$ hours.
1. Part A.
  Write the model $A(t)$ for the colony's size, $t$ hours after it starts, using the tripling-time form.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 3 points
2. Part B.
  Find the colony's size after $15$ hours.
  
  Carry your own answer forward Use your own model from part A, evaluated at $t=15$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Find exactly how many hours it takes the colony to reach $250$ cells, and confirm your answer is reasonable by checking that the colony's size at $t=8$ hours and at $t=9$ hours brackets $250$ .
  
  Carry your own answer forward Use your own model from part A, setting it equal to $250$ .
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
The answer

Part A

$A(t)=40\cdot3^{t/5}$ .

Part B

$1080$ cells.

Part C

$t\approx8.34$ hours, which lies between $8$ and $9$ since $A(8)\approx232<250<A(9)\approx289$ .

Worked solution

Part A

Since the colony triples every $5$ hours, its model takes the tripling-time form, the direct analogue of the doubling-time form with $2$ replaced by $3$ .

$A(t) = 40\cdot3^{t/5}$

Part B

Substitute $t=15$ , so the exponent is $\tfrac{15}{5}=3$ , meaning three triplings have passed.

$A(15) = 40\cdot3^{3} = 40\cdot27 = 1080$

Part C

Set the model equal to $250$ and isolate the power of $3$ .

$40\cdot3^{t/5}=250 \;\Longrightarrow\; 3^{t/5}=6.25$

Take logarithms and use the power law.

$\frac{t}{5}\log3=\log6.25 \;\Longrightarrow\; t=\frac{5\log6.25}{\log3}\approx8.34$

Check the bracket: $A(8)=40\cdot3^{1.6}\approx231.98$ and $A(9)=40\cdot3^{1.8}\approx288.99$ , and indeed $232<250<289$ , so $t\approx8.34$ falls between $8$ and $9$ hours as it should.

In one line

$A(t)=40\cdot3^{t/5}$ ; after $15$ hours the colony has $1080$ cells; and it takes about $8.34$ hours to reach $250$ cells, consistent with $A(8)\approx232$ and $A(9)\approx289$ bracketing the target.

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

Identifies the correct tripling-time form with $d=5$ . . Worth 2 points.

Uses the correct starting value of $40$ cells. . Worth 1 point.

Part B 3 points

Evaluates the exponent correctly before raising the base. . Worth 2 points.

Reports the final cell count with units. . Worth 1 point.

Part C 4 points

Divides by the starting amount and takes logarithms to solve for $t$ correctly. . Worth 2 points.

Confirms the answer is reasonable by checking that $A(8)<250<A(9)$ . . Worth 2 points.
9. Matching a continuous rate with a once-an-hour schedule . 12 points. Question 9 of 10.

A culture grows continuously. Two measurements are taken: $P(0)=60$ cells and $P(12)=90$ cells, twelve hours later. This question infers the continuous growth rate from those two readings, then compares it to the EFFECTIVE per-hour rate a once-per-hour (discrete, compounded hourly) schedule would need to match the same two readings exactly.
1. Part A.
  Use the two measurements to find the continuous growth rate $r$ (as a percent, to three decimal places), using the model $P(t)=P(0)e^{rt}$ .
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
2. Part B.
  A discrete, once-per-hour compounding schedule $D(t)=D(0)(1+i)^{t}$ ( $t$ in whole hours) is required to match the SAME two readings, $D(0)=60$ and $D(12)=90$ . Find the effective hourly rate $i$ (as a percent, to three decimal places), and state which rate is larger, $r$ or $i$ , without yet explaining why.
  
  Carry your own answer forward Compare your value of $i$ to your own value of $r$ from part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Explain, in general (not just for these two specific numbers), why the discrete effective rate $i$ needed to match a continuous rate $r$ over the SAME elapsed time must always be larger than $r$ itself, tying your answer to the inequality $e^{x} > 1+x$ for every $x>0$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$r\approx0.03379$ , about $3.379\%$ per hour.

Part B

$i\approx0.03437$ , about $3.437\%$ per hour; $i>r$ .

Part C

Matching the two schedules exactly requires $1+i=e^{r}$ (same per-hour growth factor). Since $e^{x}>1+x$ for every $x>0$ , setting $x=r$ gives $e^{r}>1+r$ , so $i=e^{r}-1>r$ : the discrete effective rate must always exceed the continuous rate.

Worked solution

Part A

$P(12)=P(0)e^{12r} \;\Longrightarrow\; 90=60e^{12r} \;\Longrightarrow\; e^{12r}=1.5 \;\Longrightarrow\; 12r=\ln1.5\approx0.405465 \;\Longrightarrow\; r\approx0.033789$

So the continuous rate is about $3.379\%$ per hour.

Part B

$D(12)=D(0)(1+i)^{12} \;\Longrightarrow\; 90=60(1+i)^{12} \;\Longrightarrow\; (1+i)^{12}=1.5 \;\Longrightarrow\; 1+i=1.5^{1/12}\approx1.034366 \;\Longrightarrow\; i\approx0.034366$

So the effective hourly rate is about $3.437\%$ , which is LARGER than the continuous rate $r\approx3.379\%$ found in part A.

Part C

For the discrete schedule to reproduce the SAME values as the continuous one at every hour, not just at $t=12$ , the two per-hour growth factors must match exactly:

$1+i = e^{r}.$

Now apply the given inequality $e^{x}>1+x$ , valid for every $x>0$ , with $x=r$ (since $r>0$ for a growing quantity):

$e^{r} > 1+r.$

Substituting $1+i=e^{r}$ into this inequality,

$1+i > 1+r \;\Longrightarrow\; i > r.$

So the discrete effective rate is always strictly larger than the continuous rate needed to match it, for ANY positive $r$ , not only the $3.379\%$ found in part A. Continuous compounding is more efficient at turning a given per-hour rate into growth, so a discrete schedule needs a boosted rate to keep pace.

In one line

The continuous rate is $r\approx3.379\%$ per hour (from $e^{12r}=1.5$ ); the equivalent discrete hourly rate is $i\approx3.437\%$ (from $(1+i)^{12}=1.5$ ), so $i>r$ ; and in general, matching requires $1+i=e^{r}$ , and since $e^{r}>1+r$ for every $r>0$ , it follows that $i=e^{r}-1>r$ always, meaning a discrete schedule always needs a higher stated rate than an equivalent continuous 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

Sets up $P(12)=P(0)e^{12r}$ and isolates $e^{12r}$ . . Worth 1 point.

Takes $\ln$ of both sides and solves for $r$ correctly. . Worth 2 points.

Reports the rate as a percent, to three decimal places. . Worth 1 point.

Part B 4 points

Sets up $(1+i)^{12}=1.5$ from the two given readings. . Worth 1 point.

Solves for $i$ correctly using the twelfth root. . Worth 2 points.

Correctly states that $i>r$ . . Worth 1 point.

Part C 4 points

Sets up the matching condition $1+i=e^{r}$ and applies the given inequality with $x=r$ . . Worth 2 points. needs an explanation, not just an answer

Concludes $i>r$ in general and connects it to continuous compounding being more efficient than discrete compounding at the same rate. . Worth 2 points.
10. A budget that multiplies against one that only grows by degree . 11 points. Question 10 of 10.

Let $f(x) = 3^{x}$ and $q(x) = x^{5}$ .
1. Part A.
  Evaluate $f$ and $q$ at $x=5$ and $x=6$ , and state which function is larger at each input.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Without computing further values, explain why $f$ must EVENTUALLY overtake $q$ permanently, even though $q$ is ahead at both inputs checked in part A, using the fact that $f$ 's per-step growth factor is fixed while $q$ 's is not.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
3. Part C.
  Now consider a second exponential, $h(x)=1.2^{x}$ , racing against the SAME polynomial $q(x)=x^{5}$ . Without evaluating any specific values of $h$ or $q$ , explain whether $h$ must ALSO eventually overtake $q$ permanently, and explain whether $h$ 's crossover point should happen sooner or later than $f$ 's crossover point found in parts A and B, using only the sizes of the two bases $1.2$ and $3$ .
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

$f(5)=243<q(5)=3125$ ; $f(6)=729<q(6)=7776$ ; $q$ leads at both.

Part B

$f$ 's per-step factor is fixed forever at $3$ , while $q$ 's per-step factor $\left(\tfrac{x+1}{x}\right)^{5}$ shrinks toward $1$ as $x$ grows; a fixed factor must eventually exceed a shrinking one, and once it does, the lead never returns.

Part C

Yes, $h$ must also eventually overtake $q$ permanently, since its base $1.2>1$ gives a fixed per-step factor forever, exactly like $f$ ; but because $1.2$ is much closer to $1$ than $3$ is, $q$ 's shrinking factor takes longer to drop below $1.2$ than below $3$ , so $h$ 's crossover happens later than $f$ 's.

Worked solution

Part A

$f(5)=3^{5}=243,\ q(5)=5^{5}=3125; \qquad f(6)=3^{6}=729,\ q(6)=6^{5}=7776$

At both inputs $q$ is larger: $243<3125$ and $729<7776$ .

Part B

Every step of $f$ multiplies by the same fixed factor $3$ , forever, since $f(x+1)/f(x)=3$ for every $x$ . The step factor of $q$ shrinks toward $1$ as $x$ grows, since $\dfrac1x$ shrinks.

$\frac{f(x+1)}{f(x)}=3 \text{ (fixed)}, \qquad \frac{q(x+1)}{q(x)}=\left(\frac{x+1}{x}\right)^{5} \longrightarrow 1 \text{ as } x\to\infty$

A fixed number greater than $1$ and a quantity shrinking toward $1$ must eventually cross, so there is some point past which $f$ 's fixed step beats $q$ 's shrinking step at every later input. Once $f$ takes the lead, every further step multiplies it by more than $q$ is multiplied by, so the gap only grows and the lead can never pass back.

Part C

Every exponential with base greater than $1$ keeps a fixed per-step growth factor forever; that part of the earlier argument has nothing to do with which particular base is racing against $q$ , so it applies to $h$ exactly as it applied to $f$ . Since $1.2>1$ , $h$ has a fixed factor of $1.2$ forever, and $q$ 's per-step factor still shrinks toward $1$ as $x$ grows, regardless of which exponential it is being compared to. A fixed factor greater than $1$ must still eventually exceed a factor shrinking toward $1$ , so $h$ permanently overtakes $q$ as well, by the identical reasoning.

The two crossovers happen at different places, though. Starting out large, $q$ 's per-step factor decreases toward $1$ as $x$ grows. It only has to shrink down past $3$ for $f$ to take the lead, but it must shrink much further, down past the smaller number $1.2$ , for $h$ to take the lead. Since $q$ 's factor is heading toward $1$ from above and $1.2$ sits much closer to $1$ than $3$ does, $q$ 's factor reaches the neighborhood of $3$ well before it reaches the neighborhood of $1.2$ . So $h$ 's crossover happens LATER than $f$ 's: a base closer to $1$ still wins eventually, but it takes longer to do so.

$q(x+1)/q(x) \to 1 \text{ from above; crosses below } 3 \text{ before it crosses below } 1.2 \quad\Longrightarrow\quad x_{f\text{-crossover}} < x_{h\text{-crossover}}$

In one line

$q$ leads at $x=5$ ( $3125>243$ ) and $x=6$ ( $7776>729$ ), but $f$ must eventually overtake $q$ permanently, because $f$ 's per-step factor stays fixed at $3$ forever while $q$ 's per-step factor shrinks toward $1$ ; a second exponential $h(x)=1.2^{x}$ must also eventually overtake $q$ permanently, by the same argument, but since $1.2$ is much closer to $1$ than $3$ is, $q$ 's shrinking per-step factor takes longer to drop below $1.2$ than below $3$ , so $h$ 's crossover happens later than $f$ 's.

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

Computes all four values correctly. . Worth 2 points.

Correctly identifies $q$ as larger at both inputs. . Worth 1 point.

Part B 4 points

States that $f$ 's per-step factor is fixed while $q$ 's per-step factor shrinks toward $1$ . . Worth 2 points.

Explains why a fixed factor eventually and permanently beats a shrinking one. . Worth 2 points. needs an explanation, not just an answer

Part C 4 points

Correctly states that $h$ also eventually overtakes $q$ permanently, using the same general argument as parts A and B. . Worth 2 points. needs an explanation, not just an answer

Correctly reasons that $h$ 's crossover happens later than $f$ 's, since $1.2$ is much closer to $1$ than $3$ is. . Worth 2 points.

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 table with one entry missing, and the base it pins down 12 points

Part A 4

Uses the ratio test on the two readings one step apart, $f(2)/f(1)$ , to find $b$ directly. 2
Checks the resulting $b$ against the wider three-step gap between $f(1)$ and $f(4)$ , and finds $a$ from $f(1)$ . 2

Part B 4

Substitutes $x=3$ into the rule found in part A to compute the missing value. 2
Verifies the missing value against both neighboring one-step ratios, confirming the table is consistent throughout. 2

Part C 4

Checks $b=-5$ directly against the equation $b^{3}=125$ and finds $(-5)^{3}=-125\neq125$ , so it fails outright. 2
States that even apart from that failure, an exponential function's base must be positive, which rules out $b=-5$ on its own. 2

2. A table that pins down a base no real exponential can have 12 points

Part A 4

Uses the ratio test to confirm a single value of $c$ fits every gap in the table. 1
Uses the law of exponents to write $c^{1}$ as the square of $c^{1/2}$ . 2
Concludes correctly that no real number squares to a negative number, so $c^{1/2}$ does not exist. 1

Part B 4

Sets up the general squaring identity $\left[d^{1/2}\right]^{2}=d$ for an arbitrary $d<0$ . 2
Concludes the contradiction holds for EVERY negative $d$ , not just $c=-3$ , and connects it to the base restriction on an exponential function. 2

Part C 4

Identifies a specific real input (any negative $x$ , such as $x=-1$ ) where $0^{x}$ breaks down. 2
Explains that this is a division-by-zero failure, a fundamentally different kind of breakdown from the missing-real-root failure in part A. 2

3. Reading one graph's ceiling off the other's floor 11 points

Part A 3

Swaps the coordinates of both given points correctly. 2
Reports the two resulting points clearly, without recomputing either one from the definition of a logarithm. 1

Part B 4

Confirms $8^{2}=64$ as an independent check of the point from part A. 1
Evaluates $\log_8\tfrac18$ correctly using the definition. 2
Reports both results as clean exact values. 1

Part C 4

Explains that reflection swaps the input and output roles of every point's coordinates. 2
Connects that swap explicitly to the general fact that an inverse function's domain equals the original's range, and vice versa. 2

4. Two expressions that look alike but need different checks 12 points

Part A 4

Identifies which cancellation law applies to each expression before simplifying. 2
Evaluates both expressions correctly. 1
Reports both results as plain numbers, with no leftover exponent notation. 1

Part B 4

Correctly classifies each expression as defined or not. 2
Ties the undefined case specifically to the negative argument of the inner logarithm. 2

Part C 4

Correctly finds the domain of $j$ is all real numbers and the domain of $k$ is $x<-3$ or $x>3$ . 2
Explains that the difference in domains traces to which cancellation law's inner piece is automatically positive. 2

5. Counting the digits of a number too large to write down 13 points

Part A 5

Applies the power rule to bring the exponent to the front. 2
Computes the numeric value correctly. 2
Reports the value as a decimal approximation consistent with the given approximation of $\log_{10}6$ . 1

Part B 4

States the digit-count rule connecting a logarithm's trapped value to a digit count. 1
Correctly reads off the digit count from the trapped logarithm. 2
Shows the distinction between $d-1$ and $d$ explicitly, rather than only stating the final digit count. 1

Part C 4

Explains that rounding answers a different question than flooring the logarithm and adding one. 2
Gives a general case, a logarithm with a decimal part below $0.5$ , where rounding would undercount the digits. 2

6. Finding the one broken step in someone else's solution 12 points

Part A 4

Correctly identifies Step 3 as the location of the error. 1
Correctly re-expands $(x+9)(x-3)$ using all four FOIL products. 2
Names the specific mistake (dropping the two cross terms) rather than only stating that Step 3 is wrong. 1

Part B 4

Solves the corrected quadratic and reports both candidate values of $x$ . 2
Tests both candidates against the domain from Step 1 and correctly classifies each. 2

Part C 4

Explains that the domain check only tests positivity and passes cleanly for $x=\sqrt{91}$ , revealing nothing. 2
Explains that a full substitution recomputes the original equation and would fail at $x=\sqrt{91}$ , catching the arithmetic error. 2

7. A quadratic hiding inside a repeated logarithm 13 points

Part A 5

Substitutes correctly, recognizing $(\log_6x)^{2}$ as $u^{2}$ . 2
Factors the quadratic and reports both values of $u$ . 2
States both values of $u$ clearly as intermediate results, distinguishing them from the values of $x$ still to be found. 1

Part B 4

Correctly undoes the logarithmic substitution for both values of $u$ . 2
Verifies both resulting values of $x$ in the original equation. 2

Part C 4

Explains the conceptual difference: $(\log_6x)^{2}$ squares the value of the logarithm, while $\log_6(x^{2})$ takes the logarithm of $x^{2}$ . 2
Produces a numeric counterexample (e.g. $x=6$ ) showing the two expressions disagree. 2

8. A colony that triples on its own schedule 10 points

Part A 3

Identifies the correct tripling-time form with $d=5$ . 2
Uses the correct starting value of $40$ cells. 1

Part B 3

Evaluates the exponent correctly before raising the base. 2
Reports the final cell count with units. 1

Part C 4

Divides by the starting amount and takes logarithms to solve for $t$ correctly. 2
Confirms the answer is reasonable by checking that $A(8)<250<A(9)$ . 2

9. Matching a continuous rate with a once-an-hour schedule 12 points

Part A 4

Sets up $P(12)=P(0)e^{12r}$ and isolates $e^{12r}$ . 1
Takes $\ln$ of both sides and solves for $r$ correctly. 2
Reports the rate as a percent, to three decimal places. 1

Part B 4

Sets up $(1+i)^{12}=1.5$ from the two given readings. 1
Solves for $i$ correctly using the twelfth root. 2
Correctly states that $i>r$ . 1

Part C 4

Sets up the matching condition $1+i=e^{r}$ and applies the given inequality with $x=r$ . 2
Concludes $i>r$ in general and connects it to continuous compounding being more efficient than discrete compounding at the same rate. 2

10. A budget that multiplies against one that only grows by degree 11 points

Part A 3

Computes all four values correctly. 2
Correctly identifies $q$ as larger at both inputs. 1

Part B 4

States that $f$ 's per-step factor is fixed while $q$ 's per-step factor shrinks toward $1$ . 2
Explains why a fixed factor eventually and permanently beats a shrinking one. 2

Part C 4

Correctly states that $h$ also eventually overtakes $q$ permanently, using the same general argument as parts A and B. 2
Correctly reasons that $h$ 's crossover happens later than $f$ 's, since $1.2$ is much closer to $1$ than $3$ is. 2

Multiple choice

Free response