Algebra 3 Quiz: Solving Exponential And Logarithmic Equations

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Solving Exponential And Logarithmic EquationsQuestion 1 of 20

Solve $2\log_5 x - \log_5 4=2$ .

\{10\}

\{-10, 10\}

\{-10\}

\{100\}

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Algebra 3 Quiz

Algebra 3 Quiz: Solving Exponential And Logarithmic Equations

Practice Solving Exponential And Logarithmic Equations in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Exponential And Logarithmic Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Solve $2\log_5 x - \log_5 4=2$ .

$\{10\}$ (correct answer)
$\{-10, 10\}$
$\{-10\}$
$\{100\}$

Explanation: Whenever you see a logarithmic equation, begin by noting the domain: since

\log_5 x

requires

x>0

, any negative solution is automatically invalid. Then solve systematically:

2\log_5 x-\log_5 4=2 \implies \log_5(x^2)-\log_5 4=2.

Combine the logs:

\log_5\left(\frac{x^2}{4}\right)=2.

Rewrite in exponential form:

\frac{x^2}{4}=5^2=25 \implies x^2=100 \implies x=\pm 10.

Because

x>0

, the valid solution is

x=10

, so the solution set is

\{10\}

. You can verify:

2\log_5 10-\log_5 4=\log_5(100/4)=\log_5 25=2

. The wrong choices expose common traps. The set

\{-10,10\}

comes from solving

x^2=100

but forgetting that

x

must be positive;

\log_5(-10)

is undefined. The singleton

\{-10\}

is worse: it keeps the invalid negative root and discards the valid positive one. The choice

\{100\}

often results from converting the right side to

\log_5 100

, but then forgetting that the left side is

\log_5(x^2)

,not

\log_5 x

; comparing

\log_5 x=\log_5 100

would mistakenly give

x=100

. The coefficient

2

must stay as an exponent until you isolate the log. As a strategy, always test each candidate in the original equation, especially when squares are involved. A domain check plus a quick plug-in will eliminate every distractor here—and catch similar logarithmic mistakes on the exam.

Question 2

What value of $x$ satisfies $\log_5(x+4) - \log_5(x-4) = 1$ ?

$-4$
$4$
$6$ (correct answer)
$24$

Explanation: The quotient rule gives

\frac{x+4}{x-4}=5

, so

x+4=5x-20

and

x=6

, which satisfies the domain requirement

x>4

. The value

24

comes from solving

4x=24

but not dividing by

4

. The value

-4

comes from writing

x+4=5x+20

with a sign error. The value

4

makes

\log_5(0)

undefined.

Question 3

Let $u=e^x$ . Use this substitution to solve $e^{2x} - 3e^x - 4 = 0$ .

$x=\ln(-1)$
$x=-1$ and $x=4$
$x=\ln 4$ and $x=\ln(-1)$
$x=\ln 4$ (correct answer)

Explanation: The substitution gives

u^2-3u-4=0

, so

u=4

u=-1

. Since

e^x

is positive for every real

x

, the value

u=-1

is rejected and only

e^x=4

survives, giving

x=\ln 4

. Reporting

x=-1

and

x=4

mistakes the

u

values for

x

values, and any answer containing

\ln(-1)

uses a logarithm of a negative number, which is undefined.

Question 4

Rewrite $2^{2x} - 2^x - 6 = 0$ as a quadratic in $2^x$ and solve for $x$ .

$x=\log_2 3$ and $x=\log_2(-2)$
$x=3$ and $x=-2$
$x=\log_2 3$ (correct answer)
There is no real solution

Explanation: With

u=2^x

the equation becomes

u^2-u-6=0

, so

u=3

u=-2

. An exponential with a positive base is always positive, so

u=-2

is rejected and

2^x=3

gives

x=\log_2 3

. Treating

3

and

-2

as the

x

values forgets to undo the substitution,

\log_2(-2)

is undefined, and a real solution does exist.

Question 5

Before solving $\log_2(x-4) + \log_2(x+2) = 4$ , what restriction does the domain place on $x$ ?

$x>-2$
$x>0$
$x\neq 4$
$x>4$ (correct answer)

Explanation: Both arguments must be positive, so

x-4>0

and

x+2>0

must hold at once; the stricter condition

x>4

controls. The condition

x>-2

satisfies only the second logarithm and leaves

\log_2(x-4)

undefined for values such as

x=0

. The condition

x>0

likewise admits values such as

x=1

that make

x-4

negative. Writing

x\neq 4

excludes a single point but still allows negative arguments.

Question 6

Rewrite $\log_3(x+1) = 2 - \log_3(x-7)$ with both logarithms on one side, then solve it.

$x=8$ (correct answer)
$x=-2$
$x=-2$ and $x=8$
There is no real solution

Explanation: Adding

\log_3(x-7)

to both sides gives

\log_3[(x+1)(x-7)]=2

, hence

x^2-6x-16=0

with roots

x=8

and

x=-2

. The domain requires

x>7

, so

x=-2

is extraneous and

x=8

is the only solution. Keeping

x=-2

ignores that it makes both arguments negative, and concluding there is no solution overlooks the valid root

x=8

Question 7

Determine the solution set of $\log_5(x-1) + \log_5(x+3) = \log_5(1-x)$ .

$x=-4$
$x=1$
$x=-4$ and $x=1$
The solution set is empty (correct answer)

Explanation: Equating arguments gives

x^2+3x-4=0

with roots

x=-4

and

x=1

, but the domain demands

x>1

from the left side and

x<1

from the right side, conditions that cannot hold together. Substituting

x=1

yields

\log_5 0

and substituting

x=-4

yields

\log_5(-5)

, so both roots are extraneous and no real solution remains.

Question 8

A student solves $\log(x-1) + \log(x-4) = 1$ , obtains $x=6$ and $x=-1$ , and reports both. Assess this work.

Both values are solutions of the original equation.
Only $x=6$ is a solution of the original equation. (correct answer)
Only $x=-1$ is a solution of the original equation.
Neither value is a solution of the original equation.

Explanation: The algebra is sound:

(x-1)(x-4)=10

gives

x^2-5x-6=0

with roots

6

and

-1

. The domain requires

x>4

, and substituting

x=-1

produces

\log(-2)

, which is undefined, so that root is extraneous. Substituting

x=6

gives

\log 5 + \log 2 = \log 10 = 1

, so the larger root is genuine and the answer is not that both or neither work.

Question 9

Write both sides of $27^x = 9^{x+1}$ as powers of $3$ , then solve.

$-2$
$1$
$2$ (correct answer)
$4$

Explanation: Rewriting gives

3^{3x}=3^{2x+2}

, so

3x=2x+2

and

x=2

; checking,

27^2=729=9^3

. The value

4

comes from solving

3x=2x+2

as if it were

x=2\cdot 2

. The value

1

comes from equating the bases

27

and

9

directly. The value

-2

comes from moving

2x

to the wrong side of the equation.

Question 10

Solve the equation $\ln(2x-1) = 0$ and verify the result lies in the domain.

$0$
$\tfrac{1}{2}$
$1$ (correct answer)
$\tfrac{e}{2}$

Explanation: Since

\ln A = 0

exactly when

A=1

, the equation gives

2x-1=1

and

x=1

, which satisfies the requirement

2x-1>0

. The value

\tfrac{1}{2}

comes from setting the argument equal to

0

, which makes the logarithm undefined. The value

\tfrac{e}{2}

comes from setting the argument equal to

e

rather than

1

, and

0

makes the argument negative.

Question 11

Express the exact solution of $7^x = 30$ using natural logarithms.

$x=\frac{\ln 7}{\ln 30}$
$x=\frac{\ln 30}{\ln 7}$ (correct answer)
$x=\ln 30 - \ln 7$
$x=\frac{30}{7}$

Explanation: Taking

\ln

of both sides gives

x\ln 7 = \ln 30

, so dividing by

\ln 7

yields

x=\frac{\ln 30}{\ln 7}

. The reciprocal

\frac{\ln 7}{\ln 30}

divides in the wrong order. The difference

\ln 30 - \ln 7

misapplies the quotient rule, which converts a logarithm of a quotient rather than a quotient of logarithms. The ratio

\frac{30}{7}

ignores the logarithms altogether.

Question 12

A colony grows according to $P = 500e^{0.04t}$ , with $t$ in years. Which expression gives the time when the colony reaches $1500$ ?

$t=\frac{\ln 3}{0.04}$ (correct answer)
$t=\frac{\ln 1500}{0.04}$
$t=\frac{3}{0.04}$
$t=0.04\ln 3$

Explanation: Dividing by

500

first gives

e^{0.04t}=3

, so

0.04t=\ln 3

and

t=\frac{\ln 3}{0.04}

. The expression

\frac{\ln 1500}{0.04}

takes the logarithm before dividing by the initial amount. The expression

\frac{3}{0.04}

never applies a logarithm. The expression

0.04\ln 3

multiplies by the rate instead of dividing by it.

Question 13

Use the difference of squares after condensing to solve $\log_2(x+3) + \log_2(x-3) = 4$ .

$x=5$ (correct answer)
$x=-5$
$x=-5$ and $x=5$
$x=\pm 25$

Explanation: Condensing gives

x^2-9=16

, so

x^2=25

and

x=\pm 5

. The domain requires

x>3

, so

x=-5

is extraneous because it makes both arguments negative, leaving

x=5

, which gives

\log_2 8 + \log_2 2 = 4

. Keeping both roots or keeping only the negative root skips that check, and

x=\pm 25

forgets to take the square root of

25

Question 14

Condense the right side of $\log(2x+1) = \log x + \log 3$ and find $x$ .

$\tfrac{1}{5}$
$1$ (correct answer)
$2$
$3$

Explanation: The product rule gives

\log(2x+1)=\log(3x)

, so

2x+1=3x

and

x=1

, which keeps both arguments positive. The value

2

comes from writing

2x+1=x+3

, adding the logarithms' arguments instead of multiplying them. The value

3

comes from reading

\log 3

as the answer itself. The value

\tfrac{1}{5}

comes from solving

2x+1=3-x

after a sign error.

Question 15

The equation $\log_2 x + \log_2(x+3) = 2$ yields the candidates $x=1$ and $x=-4$ . Why is $x=-4$ not a solution?

It makes the argument of a logarithm negative, which is undefined. (correct answer)
It makes the left side of the equation equal zero rather than two.
Negative numbers can never be solutions of any logarithmic equation.
It fails to satisfy the quadratic $x^2+3x-4=0$ that condensing produced.

Explanation: Substituting

x=-4

gives

\log_2(-4)

and

\log_2(-1)

, and logarithms are defined only for positive arguments, so the candidate is extraneous. It does satisfy

x^2+3x-4=0

, which is why the algebra produced it in the first place, so the claim that it fails the quadratic is false. The left side is undefined rather than equal to zero. A negative value can solve a logarithmic equation such as

\log_2(x+8)=2

, so the blanket claim about negative numbers is also false.

Question 16

A sample decays by $A = A_0\left(\tfrac{1}{2}\right)^{t/12}$ , with $t$ in hours. After how many hours does one eighth of the sample remain?

$4$
$12$
$24$
$36$ (correct answer)

Explanation: Setting

\left(\tfrac{1}{2}\right)^{t/12}=\tfrac{1}{8}=\left(\tfrac{1}{2}\right)^3

gives

\tfrac{t}{12}=3

and

t=36

, which is three half-lives. The value

12

is one half-life and leaves half the sample. The value

24

is two half-lives and leaves a quarter. The value

4

divides

12

3

instead of multiplying.

Question 17

What is the solution set of $\log_2 x + \log_2(x-2)=3$ ?

$\{4\}$ (correct answer)
$\{4,-2\}$
$\{-2\}$
$\{5\}$

Explanation: When you see a logarithmic equation, your first instinct should be to combine logs using the product rule, but also remember: logarithms are only defined for positive inputs. That domain check is the hidden key here. Combine the logs:

\log_2 x+\log_2(x-2)=\log_2\bigl(x(x-2)\bigr)=3

x(x-2)=2^3=8 \quad\Rightarrow\quad x^2-2x-8=0

(x-4)(x+2)=0

giving

x=4

x=-2

. However, the original equation requires

x>0

and

x-2>0

, so

x>2

. Thus

-2

is extraneous, and the solution set is

\{4\}

. The choice

\{4,-2\}

includes the algebraic root but forgets to check the logarithm domain, which is the classic trap. The choice

\{-2\}

goes further and keeps only the invalid root; it may come from solving the quadratic but ignoring the domain requirement entirely. The choice

\{5\}

does not satisfy the equation:

\log_2 5+\log_2 3\approx 3.91

, not

3

, so it is simply a guess or a substitution error. Study tip: for any logarithmic equation, find the domain before you solve, then reject any candidate that falls outside it. A solution that works algebraically but not in the original equation is extraneous — always test your roots.

Question 18

Solve $\log_3 x + \log_3(x-6) = 3$ over the real numbers.

$x=-3$ only
$x=9$ only (correct answer)
$x=-3$ and $x=9$
There is no real solution

Explanation: Condensing gives

x(x-6)=27

, so

x^2-6x-27=0

and

x=9

x=-3

. The domain requires

x>0

and

x>6

, so

x=-3

is extraneous and must be discarded, while

x=9

checks. Keeping both values ignores that

\log_3(-3)

is undefined, and rejecting both overlooks that

x=9

satisfies the original equation.

Question 19

Apply the power rule first, then solve $2\log_3 x = \log_3(2x+3)$ .

$x=-1$
$x=3$ (correct answer)
$x=-1$ and $x=3$
There is no real solution

Explanation: The power rule gives

\log_3 x^2 = \log_3(2x+3)

, so

x^2-2x-3=0

with roots

x=3

and

x=-1

. The left side requires

x>0

, so

x=-1

is extraneous because

\log_3(-1)

is undefined, while

x=3

gives

2\log_3 3 = \log_3 9

. Keeping both roots skips that check, and rejecting both discards the valid root

x=3

Question 20

Condense the left side of $\log_2(x+6) - \log_2 x = \log_2 x$ , then find every solution.

$x=-2$
$x=-2$ and $x=3$
$x=3$ (correct answer)
There is no real solution

Explanation: The quotient rule gives

\frac{x+6}{x}=x

, so

x^2-x-6=0

with roots

x=3

and

x=-2

. Every logarithm here requires

x>0

, so

x=-2

is extraneous, while

x=3

gives

\log_2 9 - \log_2 3 = \log_2 3

. Keeping both roots ignores the domain, and rejecting both overlooks the root that checks.