Exponential and Logarithmic Functions: 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.

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Question 1 of 20

1

Evaluate $\log_6 1296$ .

Answer choices for question 1 A $4$ B $3$ C $6$ D $1296$
2

Which of the following values of $b$ is an allowed base for an exponential function $f(x)=a\cdot b^{x}$ ?

Answer choices for question 2 A $b=-6$ B $b=1$ C $b=0.6$ D $b=0$
3

A river otter population declines by $9\%$ each year. Which model gives its population $t$ years from now, starting from a population of $P$ ?

Answer choices for question 3 A $P(0.09)^{t}$ B $P(0.91)^{t}$ C $P(1.09)^{t}$ D $P(1-0.09t)$
4

Which expression is equal to $\log_7\!\left(\dfrac{49x^{5}}{y^{3}}\right)$ for $x,y>0$ ?

Answer choices for question 4 A $49+5\log_7x-3\log_7y$ B $2+5\log_7x-3y$ C $2+\log_7(5x)-\log_7(3y)$ D $2+5\log_7x-3\log_7y$
5

An exponential function $f(x)=a\cdot b^{x}$ satisfies $f(3)=54$ and $f(6)=1458$ . What is $b$ ?

Answer choices for question 5 A $3$ B $27$ C $9$ D $3\sqrt{3}$
6

What is the range of $g(x) = 3\cdot5^{\,x+2} - 20$ ?

Answer choices for question 6 A $y>-20$ B $y>20$ C $y<-20$ D $y>55$
7

What is the domain of $y=\log_4\!\left(x^{2}-16\right)$ ?

Answer choices for question 7 A $x>4$ B $x<-4$ or $x>4$ C $-4<x<4$ D $x\neq\pm4$
8

Evaluate $\ln\!\left(e^{9}\right) - \ln 1$ .

Answer choices for question 8 A $0$ B $e^{9}$ C $9$ D $8$
9

Solve $25^{\,x-2}=125^{\,x+1}$ .

Answer choices for question 9 A $x=7$ B $x=-4$ C $x=-5$ D $x=-7$
10

$P$ dollars grows at an annual rate of $9\%$ compounded semiannually for $7$ years. Which expression gives the balance?

Answer choices for question 10 A $P(1.045)^{14}$ B $P(1.09)^{14}$ C $P(1.045)^{7}$ D $P(1.18)^{7}$
11

What is $\log_{16} 8$ ?

Answer choices for question 11 A $\tfrac43$ B $\tfrac34$ C $2$ D $\tfrac12$
12

The graph of $y=6^{x}$ passes through $\left(-1,\tfrac16\right)$ and $(3,216)$ . Which two points must lie on the graph of $y=\log_6 x$ ?

Answer choices for question 12 A $\left(-1,\tfrac16\right)$ and $(3,216)$ B $\left(\tfrac16,1\right)$ and $(216,-3)$ C $\left(\tfrac16,-1\right)$ and $(216,3)$ D $\left(-\tfrac16,-1\right)$ and $(-216,3)$
13

What is the complete solution set of $\log_3 x + \log_3(x-8) = 2$ ?

Answer choices for question 13 A $\{9,-1\}$ B $\{-1\}$ C No solution. D $\{9\}$
14

An exponential function $f(x)=a\cdot b^{x}$ satisfies $f(1)=6$ and $f(3)=54$ . Compared with the polynomial $g(x)=x^{4}$ , which statement must be true for large enough $x$ ?

Answer choices for question 14 A $f(x)$ eventually exceeds $g(x)$ and stays ahead forever. B $g(x)$ eventually exceeds $f(x)$ and stays ahead forever, since a fixed base cannot keep up with a growing polynomial exponent. C $f(x)$ and $g(x)$ cross exactly once, and $g(x)$ leads afterward. D Neither ever overtakes the other permanently, since the base found, $b=3$ , is too small.
15

Evaluate $27^{\log_3 4}$ .

Answer choices for question 15 A $4$ B $64$ C $12$ D $81$
16

Which expression correctly models $1500$ dollars growing continuously for $9$ years at an annual rate of $3.5\%$ ?

Answer choices for question 16 A $1500(1.035)^{9}$ B $1500e^{31.5}$ C $1500e^{0.315}$ D $1500e^{0.035/9}$
17

The point $(a,4)$ lies on the graph of $y=\log_2 x$ . Using the definition of a logarithm, find $a$ , then state the point that must lie on the graph of $y=2^{x}$ by the mirror relationship between the two graphs.

Answer choices for question 17 A $a=2$ ; the point $(4,2)$ lies on $y=2^{x}$ . B $a=8$ ; the point $(4,8)$ lies on $y=2^{x}$ . C $a=16$ ; the point $(16,4)$ lies on $y=2^{x}$ . D $a=16$ ; the point $(4,16)$ lies on $y=2^{x}$ .
18

Solve $\log_2(x+3) - \log_2(x-3) = 4$ .

Answer choices for question 18 A $x=\tfrac{17}{5}$ B $x=-\tfrac{17}{5}$ C $x=\tfrac{45}{17}$ D No solution, since the domain $x>3$ can never be satisfied.
19

A quantity grows continuously at an annual rate of $5.5\%$ . Compared to the SAME nominal rate compounded annually (once a year, not continuously), which statement about their doubling times is true?

Answer choices for question 19 A They have exactly the same doubling time, since the rate is the same number in both models. B Continuous compounding reaches double sooner than annual compounding at the same rate. C Annual compounding reaches double sooner, since compounding less often lets each period keep more of its own interest. D It cannot be determined without knowing the starting amount.
20

What is the exact solution set of $49^{x} - 4\cdot7^{x} - 45 = 0$ ?

Answer choices for question 20 A $\{\log_7 9,\ \log_7(-5)\}$ B $\{9\}$ C $\{\log_7 9\}$ D $\{\log_7 5\}$

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Where you stand

Build or identify an exponential function $f(x)=a\cdot b^{x}$ from a table of values or two points using the constant-ratio test, and explain why the base must be positive and different from $1$ . Review: Exponential Functions and Graphs
Read the transformations of an exponential or logarithmic graph (shifts, stretches, the asymptote) and describe how the two graphs mirror each other across the line $y=x$ . Review: Exponential Functions and Graphs , Introduction to Logarithms
Evaluate a logarithm directly from its definition, identify the domain restriction on its argument, and apply the two cancellation laws. Review: Introduction to Logarithms
Apply the product, quotient, and power rules for logarithms to expand or condense an expression, and convert between bases with the change-of-base formula. Review: Properties of Logarithms
Solve an exponential or logarithmic equation using the one-to-one property or by taking a logarithm of both sides, testing every candidate for an extraneous root, including an equation that hides a quadratic. Review: Solving Exponential and Logarithmic Equations
Build a constant-percent growth or decay model, apply compounding $n$ times a year, and find a doubling time or half-life. Review: Exponential Growth and Decay
Explain where the number $e$ comes from as the ceiling of ever-finer compounding, apply the continuous growth or decay model with the natural logarithm, and compare it with a discrete compounding schedule. Review: The Natural Base e
Compare exponential growth against polynomial growth, explaining why an exponential with base greater than $1$ eventually and permanently overtakes any polynomial, regardless of its degree. Review: Exponential Functions and Graphs

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)

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Problem 1 of 10

Problem 1 A nested reading

Evaluate $\log_6(6^{(\log_{17}289)^2-7})$ .
1. Hint 1
  Interpret the inner logarithm before carrying out the operations in the exponent.
2. Hint 2
  After the exponent is evaluated, the outer logarithm reverses the power with the matching base.
Answer

$-3$ .

Full solution

The inner logarithm is $\log_{17}289=2$ , since $17^2=289$ .
Its square minus seven is $2^2-7=-3$ .

The outer argument is therefore $6^{-3}$ , which is positive.
The inverse law gives
$\log_6(6^{-3})=-3$
This is also the definition of the logarithm as the exponent of base six.

Answer

$-3$ .

Key idea

Nested expressions can be evaluated from their inner operations before applying the outer inverse law.
Problem 1: Mark complete
Problem 2 One compact expression

For $x>1$ , write $\frac{\log_2(x+1)-\log_2(x-1)}{\log_2 5}$ as one logarithm with coefficient one.
1. Hint 1
  The difference in the numerator can become the logarithm of a positive quotient.
2. Hint 2
  The remaining denominator identifies a change of logarithm base.
Answer

$\log_5(\frac{x+1}{x-1})$ .

Full solution

Since $x>1$ , both $x+1$ and $x-1$ are positive.
By the quotient rule, the difference of their logarithms is $\log_2((x+1)/(x-1))$ .

The denominator $\log_2 5$ is nonzero.
Changing the base gives
$\log_5\left(\frac{x+1}{x-1}\right)$
Expanding and converting back to base $2$ recovers the original expression.

Answer

$\log_5(\frac{x+1}{x-1})$ .

Key idea

The quotient rule and change of base can combine in a single logarithmic simplification.
Problem 2: Mark complete
Problem 3 Shuffled outputs

A decreasing exponential function $f(x)=ab^x$ takes the values $-55$ , $-1375$ and $-11$ at the inputs $0$ , $1$ and $3$ , in some order. Match each output to its input, find $f$ , and explain why its negative outputs do not require a negative base.
1. Hint 1
  A decreasing function gives smaller outputs at larger inputs.
2. Hint 2
  Use the outputs at inputs 0 and 1 to find the coefficient and the base, then check the third output.
Answer

$f(0)=-11$ , $f(1)=-55$ , $f(3)=-1375$ ; $f(x)=-11\cdot5^x$ .

Full solution

Since $f$ is decreasing, the largest output belongs to the smallest input.
As $-11>-55>-1375$ , this gives $f(0)=-11$ , $f(1)=-55$ and
$f(3)=-1375.$

Then $a=f(0)=-11$ and $b=\frac{f(1)}{f(0)}$ , which is
$\frac{-55}{-11}=5.$
Checking the third output,
$-11\cdot5^3=-1375$
so $f(x)=-11\cdot5^x$ fits all three.

Every power $5^x$ is positive, so the sign of each output comes from $a=-11$ .
A negative base is not allowed, since for example $(-5)^{1/2}$ is not a real number, and the base $5$ also differs from $1$ , which would give a constant function.

Answer

$f(0)=-11$ , $f(1)=-55$ , $f(3)=-1375$ ; $f(x)=-11\cdot5^x$ .

Key idea

The sign of an exponential function's outputs can come from its coefficient, while its base stays positive and different from 1.
Problem 3: Mark complete
Problem 4 Two constructions

The graph shows $f(x)=2^x-2$ . Construction A reflects this graph across the line $y=x$ , then moves the result up $9$ units. Construction B moves the graph of $f$ up $9$ units, then reflects the result across $y=x$ . For each, give the resulting equation, its domain and its vertical asymptote. Do the two constructions give the same graph?

The graph of f.

Text description of this figure

A grid with the x-axis and y-axis each from -4 to 8 on equal scales, a grid line and a label at every integer. A dashed horizontal line runs across the window at height -2, with no label. A solid curve labeled f starts at the left edge just above that dashed line, rises slowly, crosses the y-axis at -1 and the x-axis at 1, then climbs steeply and leaves the top of the window a little past x equals 3. A dotted diagonal line through the origin, labeled y equals x, runs from the lower left corner to the upper right corner. No points are marked and no inverse is drawn.
1. Hint 1
  Reflecting across $y=x$ exchanges the roles of $x$ and $y$ .
2. Hint 2
  Carry out the steps in the order given, tracking the asymptote through each one.
Answer

A: $y=\log_2(x+2)+9$ , domain $x>-2$ , asymptote $x=-2$ . B: $y=\log_2(x-7)$ , domain $x>7$ , asymptote $x=7$ . The graphs are different.

Full solution

A: exchanging $x$ and $y$ in $y=2^x-2$ gives $x+2=2^y$ , so the reflection is $y=\log_2(x+2)$ , with domain $x>-2$ and vertical asymptote $x=-2$ , the image of $y=-2$ .
Moving up $9$ units changes neither, giving
$y=\log_2(x+2)+9$

B: moving up first gives $y=2^x+7$ , whose asymptote is $y=7$ .
Exchanging $x$ and $y$ gives $x-7=2^y$ , so
$y=\log_2(x-7)$
with domain $x>7$ and vertical asymptote $x=7$ .

The domains differ, so the graphs differ.
A vertical move made before the reflection becomes a horizontal move after it, which is why the order matters.

Answer

A: $y=\log_2(x+2)+9$ , domain $x>-2$ , asymptote $x=-2$ . B: $y=\log_2(x-7)$ , domain $x>7$ , asymptote $x=7$ . The graphs are different.

Key idea

Reflection across y = x turns a vertical shift into a horizontal one, so its place in a sequence of transformations can change the result.
Problem 4: Mark complete
Problem 5 Two savings schedules

A positive deposit must grow to $1.7$ times its starting value in $8$ years. One plan compounds continuously at annual rate $r_C$ ; the other compounds quarterly at nominal annual rate $r_Q$ . Give each rate exactly and as a percent to the nearest hundredth of a percent. Which plan needs the higher nominal rate, and why?
1. Hint 1
  The starting deposit cancels from both target equations.
2. Hint 2
  Continuous growth gives the factor $e^{8r_C}$ , while quarterly compounding over $8$ years uses $32$ equal periods.
Answer

$r_C=\frac{\ln1.7}8\approx6.63\%$ and $r_Q=4(1.7^{1/32}-1)\approx6.69\%$ ; the quarterly plan needs the higher rate.

Full solution

The deposit cancels from each target equation.
For continuous growth $e^{8r_C}=1.7$ , so
$r_C=\frac{\ln1.7}8$
which is about $0.0663$ , or $6.63\%$ .

Quarterly compounding over $8$ years uses $32$ periods, so
$(1+\frac{r_Q}4)^{32}=1.7.$
The positive period factor is $1.7^{1/32}$ , hence $r_Q=4(1.7^{1/32}-1)\approx0.0669$ , about $6.69\%$ .

At one positive nominal rate $r$ , the yearly factors $(1+\frac rn)^n$ increase with the number $n$ of periods toward the continuous factor $e^r$ , the same limit that defines $e$ .
So quarterly compounding grows more slowly than continuous compounding at the same rate, and needs a slightly higher nominal rate to reach the same target.

Answer

$r_C=\frac{\ln1.7}8\approx6.63\%$ and $r_Q=4(1.7^{1/32}-1)\approx6.69\%$ ; the quarterly plan needs the higher rate.

Key idea

For the same growth factor over the same positive time, a finite compounding schedule needs a higher positive nominal rate than continuous compounding.
Problem 5: Mark complete
Problem 6 A logarithm hiding a quadratic

Find every real solution of $\log_3(9^x+18)=x+2$ .
1. Hint 1
  Rewrite the logarithmic equation in exponential form.
2. Hint 2
  Express $9^x$ and $3^{x+2}$ through the single quantity $u=3^x$ .
Answer

$x=1$ or $x=\log_3 6$ .

Full solution

The argument $9^x+18$ is positive for every real $x$ , so no input is excluded.
By the definition of the logarithm,
$9^x+18=3^{x+2}$

With $u=3^x>0$ , $9^x=u^2$ and $3^{x+2}=9u$ , so $u^2-9u+18=0$ , which factors as
$(u-3)(u-6)=0$
Both roots are positive, so both are possible values of $3^x$ .

From $3^x=3$ , $x=1$ ; from $3^x=6$ ,
$x=\log_3 6.$
Check: at $x=1$ the argument is $9+18$ , which is $3^3$ , so the left side is $3$ , and so is $1+2$ .
At $x=\log_3 6$ , $9^x=36$ , so the argument is $54=27\cdot2$ and the left side is $3+\log_3 2$ ; the right side $\log_3 6+2$ is $1+\log_3 2+2$ , the same number.

Answer

$x=1$ or $x=\log_3 6$ .

Key idea

Rewriting a logarithmic equation in exponential form can reveal a quadratic in a positive exponential quantity, and each positive root must still be checked in the original.
Problem 6: Mark complete
Problem 7 Three readings

A positive quantity decays exponentially according to $A(t)=ab^t$ for $t\ge0$ , in hours. Its readings at $0$ , $1$ and $2$ hours total $186$ units, and the reading at $2$ hours is $\frac{16}{49}$ of the reading at $0$ . Find the model and all three readings, and give the half-life exactly and to the nearest tenth of an hour.
1. Hint 1
  Over equal time steps an exponential multiplies by the same factor.
2. Hint 2
  The two-hour ratio is the square of the hourly factor; then use the total to find the starting reading.
Answer

$A(t)=98(\frac47)^t$ ; readings $98$ , $56$ and $32$ units; half-life $\frac{\ln(1/2)}{\ln(4/7)}\approx1.2$ hours.

Full solution

The two-hour ratio is
$b^2=\frac{16}{49}.$
The base is positive, so $b=\frac47$ , which is less than $1$ , as decay requires.

The readings are $a$ , $\frac47a$ and $\frac{16}{49}a$ , so
$\frac{93}{49}a=186$
and $a=98$ .
The readings are $98$ , $56$ and $32$ , which total $186$ .

The half-life $h$ satisfies $(\frac47)^h=\frac12$ , so
$h=\frac{\ln(1/2)}{\ln(4/7)}$
which is about $1.24$ , or $1.2$ hours to the nearest tenth.
Both logarithms are negative, so $h$ is positive.

Answer

$A(t)=98(\frac47)^t$ ; readings $98$ , $56$ and $32$ units; half-life $\frac{\ln(1/2)}{\ln(4/7)}\approx1.2$ hours.

Key idea

For readings at known, distinct times, their ratio determines the positive base of an exponential model, and their total then determines its coefficient.
Problem 7: Mark complete
Problem 8 A proposed identity

For $u,v>0$ , a student claims $\ln(\frac{u+v}{uv})=\ln(u+v)-\ln u-\ln v$ . Is the claim valid for all the stated inputs? Explain.
1. Hint 1
  The numerator is one positive expression; the denominator is a positive product.
2. Hint 2
  Apply the quotient rule and then split the logarithm of the denominator.
Answer

Yes; it is valid for all $u,v>0$ .

Full solution

The numerator and denominator are positive, and $uv\ne0$ .
The quotient rule gives $\ln(u+v)-\ln(uv)$ .
Splitting the denominator's product gives
$\ln(uv)=\ln u+\ln v$
and subtracting this whole sum yields the claimed expression.
The numerator's sum remains intact.

Answer

Yes; it is valid for all $u,v>0$ .

Key idea

A sum can remain inside one logarithm while products and quotients elsewhere are expanded.
Problem 8: Mark complete
Problem 9 An argument comparison

Find all real solutions of $\log_2(x^2-1)=2\log_2(x-1)$ , and justify whether each candidate belongs to the original domain.
1. Hint 1
  The right logarithm restricts $x$ before any rewriting.
2. Hint 2
  On that domain, the right side is the logarithm of $(x-1)^2$ .
Answer

No real solution; the candidate $x=1$ is outside the domain.

Full solution

The right argument requires $x>1$ , which also makes $x^2-1>0$ .
The equation then gives $x^2-1=(x-1)^2$ , so
$2x=2$
and the sole candidate is $x=1$ .

At that value both logarithm arguments are zero.
It is not in the domain, so there is no real solution.

Answer

No real solution; the candidate $x=1$ is outside the domain.

Key idea

A candidate at a logarithm's domain boundary must be rejected even if it solves the transformed algebra.
Problem 9: Mark complete
Problem 10 A late overtaking

For positive integers $n$ , compare $E(n)=2^n$ with $P(n)=n^{12}$ . Using exponent laws rather than a calculator, decide which is larger at $n=64$ and at $n=128$ . Then explain why $E(n)>P(n)$ for every $n\ge128$ .
1. Hint 1
  Write $64$ and $128$ as powers of $2$ .
2. Hint 2
  Compare the factor by which each function grows from $n$ to $n+1$ .
Answer

At $n=64$ , $P$ is larger: $2^{72}>2^{64}$ . At $n=128$ , $E$ is larger: $2^{128}>2^{84}$ . $E(n)>P(n)$ for every integer $n\ge128$ .

Full solution

Since $64=2^6$ , $P(64)=(2^6)^{12}=2^{72}$ , which exceeds
$E(64)=2^{64}.$
Since $128=2^7$ , $P(128)=2^{84}$ , which is less than
$E(128)=2^{128}.$

From $n$ to $n+1$ , $E$ is multiplied by $2$ and $P$ by
$\frac{P(n+1)}{P(n)}=\left(1+\frac1n\right)^{12}$
Since $1+\frac1n$ shrinks as $n$ grows, for $n\ge128$ this factor is at most $(1+\frac1{128})^{12}\approx1.10$ , less than $2$ .

So for $n\ge128$ the ratio $E(n)/P(n)$ is multiplied by more than $1$ at every step.
It is greater than $1$ at $n=128$ , so it stays greater than $1$ , and $E(n)>P(n)$ for every $n\ge128$ .

Answer

At $n=64$ , $P$ is larger: $2^{72}>2^{64}$ . At $n=128$ , $E$ is larger: $2^{128}>2^{84}$ . $E(n)>P(n)$ for every integer $n\ge128$ .

Key idea

When an exponential and a polynomial both stay positive, once the exponential is ahead and its fixed step factor beats the polynomial's from then on, it stays ahead for good.
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Algebra II. You can skip it. → Continue to Chapter 11: Sequences and Series →

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