Algebra 3 Quiz: Logarithm Properties

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Logarithm PropertiesQuestion 1 of 12

Which expression is equivalent to $\log_5(25x^3) - 2\log_5(5x)$ for $x>0$ ?

\log_5(5x)

\log_5 x

2\log_5 x

1

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Algebra 3 Quiz

Algebra 3 Quiz: Logarithm Properties

Practice Logarithm Properties 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 Logarithm Properties, 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

Which expression is equivalent to $\log_5(25x^3) - 2\log_5(5x)$ for $x>0$ ?

$\log_5(5x)$
$\log_5 x$ (correct answer)
$2\log_5 x$
$1$

Explanation: When you see a logarithm expression with coefficients and products, your first move should be to expand using the product and power rules:

\log_b(MN)=\log_b M+\log_b N

and

\log_b(M^a)=a\log_b M

. Here,

\log_5(25x^3)=\log_5 25+\log_5(x^3)=2+3\log_5 x

. Also,

2\log_5(5x)=2(\log_5 5+\log_5 x)=2(1+\log_5 x)=2+2\log_5 x

. Subtracting gives

(2+3\log_5 x)-(2+2\log_5 x)=\log_5 x

, so the equivalent expression is

\log_5 x

. The choice

\log_5(5x)

is a trap from moving the coefficient 2 inside incorrectly:

2\log_5(5x)

means

\log_5((5x)^2)=\log_5(25x^2)

, not

\log_5(5x^2)

. The choice

2\log_5 x

comes from distributing the 2 only to part of the log:

2\log_5(5x)

must become

2+2\log_5 x

, not

2+\log_5 x

. The choice

1

treats

\log_5(5x)

as if it were just

\log_5 5=1

, forgetting the

x

inside; actually

\log_5(5x)=1+\log_5 x

. A good check is to plug in a non-special value like

x=2

. Avoid testing

x=5

, because

\log_5 5=1

would make the constant

1

look falsely correct. Always expand fully, and make sure coefficients apply to the entire argument inside the log.

Question 2

Which expression is equivalent to $2\log_3 x - \log_3(x+1) + \log_3(x-1)$ for $x>1$ ?

$\log_3\left(\frac{x^2(x+1)}{x-1}\right)$
$\log_3\left(\frac{x^2}{x^2-1}\right)$
$\log_3\left(\frac{2x(x-1)}{x+1}\right)$
$\log_3\left(\frac{x^2(x-1)}{x+1}\right)$ (correct answer)

Explanation: When you see logarithms being added and subtracted with coefficients, your first thought should be the product, quotient, and power properties. Combine the terms into a single logarithm, being careful that subtraction means the quotient and that a coefficient becomes an exponent. Start with the coefficient:

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

. Then handle the difference and sum of the remaining logs:

-\log_3(x+1)+\log_3(x-1) = \log_3\left(\frac{x-1}{x+1}\right)

. Now multiply the two results inside one log:

\log_3(x^2) + \log_3\left(\frac{x-1}{x+1}\right) = \log_3\left(\frac{x^2(x-1)}{x+1}\right)

. That is the correct expression. The choice

\log_3\left(\frac{x^2(x+1)}{x-1}\right)

flips the quotient, turning

(x-1)/(x+1)

into

(x+1)/(x-1)

— a sign-order error. The choice

\log_3\left(\frac{x^2}{x^2-1}\right)

shows the common mistake of writing

x^2/(x^2-1)

instead of

x^2(x-1)/(x+1)

; these are not equivalent because the

x-1

belongs in the numerator, not as part of a factored denominator. Finally,

\log_3\left(\frac{2x(x-1)}{x+1}\right)

confuses

2\log_3 x

with

\log_3(2x)

— but the power property says the coefficient becomes an exponent, so

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

, not

\log_3(2x)

. A reliable study tip: before combining logs, rewrite every coefficient as an exponent, and always test with a simple value like

x=2

to verify equivalence. If the two expressions give the same number, you've likely combined them correctly.

Question 3

Which expression is equivalent to $\log_b\left(\frac{1}{b\sqrt[3]{x}}\right)$ for $x>0$ ?

$1-\frac{1}{3}\log_b x$
$-1-\frac{1}{3}\log_b x$ (correct answer)
$-1+\frac{1}{3}\log_b x$
$-\frac{1}{3}\log_b x$

Explanation: Whenever you see a logarithm of a quotient or a root, your first move should be to unpack it using logarithm laws:

\log(M/N)=\log M-\log N

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

, and

\log(M^k)=k\log M

. Here,

\log_b\left(\frac{1}{b\sqrt[3]{x}}\right)

is a quotient whose denominator is

b x^{1/3}

, so rewrite it as

\log_b 1 - \log_b(b x^{1/3})

. Since

\log_b 1=0

, this becomes

-\left(\log_b b + \log_b x^{1/3}\right)

. Now

\log_b b=1

, and

\log_b x^{1/3}=\frac13\log_b x

. Therefore the expression equals

-1-\frac13\log_b x

. The option

-1+\frac13\log_b x

gets the

-1

correct but flips the sign on the logarithm term; it treats

\log_b(1/\sqrt[3]{x}) = \frac13\log_b x

, but the log of a reciprocal is negative. The option

1-\frac13\log_b x

has both sign errors; it would correspond to

\log_b(b\sqrt[3]{x})

, not the quotient with denominator

b\sqrt[3]{x}

. The option

-\frac13\log_b x

omits the

-1

entirely, forgetting that the

b

in the denominator contributes

\log_b b=1

. When you see a fraction inside a logarithm, expand carefully in steps: simplify

\log(1/\text{stuff})

-\log(\text{stuff})

, then apply product and power rules. Watch every sign, especially the leading negative, and remember that the base

b

itself becomes a constant logarithm term like

1

Question 4

Which expression is NOT equivalent to $\log_2\left(\frac{16x^2}{y}\right)$ for $x,y>0$ ?

$4+2\log_2 x - \log_2 y$
$2\log_2(4x) - \log_2 y$
$\log_2(16x^2) - \log_2 y$
$4+\log_2 x^2 - \log_2(2y)$ (correct answer)

Explanation: Whenever you see a logarithmic expression with products, quotients, and powers, your first move should be to expand it using the log laws:

\log_b(MN)=\log_b M+\log_b N

\log_b(M/N)=\log_b M-\log_b N

, and

\log_b(M^k)=k\log_b M

. Starting from

\log_2\left(\frac{16x^2}{y}\right)

, split the quotient:

\log_2(16x^2)-\log_2 y

, then the product and power:

\log_2 16+2\log_2 x-\log_2 y=4+2\log_2 x-\log_2 y

. That is the target form. The choice

4+2\log_2 x-\log_2 y

is exactly this expansion, so it is equivalent. The choice

\log_2(16x^2)-\log_2 y

is just the first quotient-splitting step, also equivalent. The choice

2\log_2(4x)-\log_2 y

works because

2\log_2(4x)=2(\log_2 4+\log_2 x)=2(2+\log_2 x)=4+2\log_2 x

. All three simplify to the original. The non-equivalent expression is

4+\log_2 x^2-\log_2(2y)

. It looks close, but

\log_2(2y)=\log_2 2+\log_2 y=1+\log_2 y

, so it becomes

4+2\log_2 x-(1+\log_2 y)=3+2\log_2 x-\log_2 y

, missing 1. That extra factor of 2 inside the logarithm becomes

+1

when pulled out, and subtracting it changes the constant term. Study tip: when comparing logarithmic forms, expand every choice completely into the same base terms. Watch for coefficients hidden inside logs, like

\log_2(2y)

, which secretly contribute a constant.

Question 5

Which value is equivalent to $\log_4 27 \cdot \log_9 8$ ?

$\frac{9}{4}$ (correct answer)
$\frac{3}{2}$
$2$
$3$

Explanation: When you see a product of logarithms with different bases, your first thought should be: can I rewrite the bases and arguments as powers of the same numbers? Here,

27=3^3

8=2^3

4=2^2

, and

9=3^2

. Using the power rule

\log_{a^m} b^n = \frac{n}{m}\log_a b

, we get

\log_4 27 = \log_{2^2}3^3 = \frac{3}{2}\log_2 3

, and

\log_9 8 = \log_{3^2}2^3 = \frac{3}{2}\log_3 2

. Multiplying gives

\frac{9}{4}(\log_2 3)(\log_3 2)

. Since

\log_a b \cdot \log_b a = 1

, the product simplifies to

\frac{9}{4}

. That is the correct value. The choice

\frac{3}{2}

is a trap for forgetting the reciprocal identity — you might correctly compute each factor as

\frac{3}{2}\log_2 3

and

\frac{3}{2}\log_3 2

, then mistakenly multiply only the coefficients but leave the log product as is, or think the log product cancels to zero. The choice

2

might come from misapplying the change of base formula or incorrectly simplifying

\log_4 27

\frac{3}{2}

and

\log_9 8

\frac{3}{2}

then adding instead of multiplying. The choice

3

could arise from treating

\log_4 27

3\log_4 3

and

\log_9 8

3\log_9 2

, then incorrectly assuming

\log_4 3 \cdot \log_9 2 = 1

without changing bases. Your takeaway: whenever you see

\log_a b \cdot \log_b c

, consider using

\log_a b = \frac{1}{\log_b a}

to simplify the product to

\log_a c

. Here, the key was recognizing that

27

and

8

are powers of

3

and

2

, allowing you to factor out the exponents cleanly. Practice rewriting logs with matching base-argument pairs to spot these cancellations quickly.

Question 6

Solve $\log_2(x+1)+\log_2(x-1)=3$ . Which value of $x$ satisfies the equation?

$\pm 3$
$-3$
$3$ (correct answer)
$9$

Explanation: When you see a sum of logs with the same base, your first move should be to combine them using

\log_b M+\log_b N=\log_b(MN)

. But always keep in mind that a logarithm is only defined when its input is positive—this domain check is what decides between otherwise valid-looking answers. Here, combine the logs:

\log_2(x+1)+\log_2(x-1)=\log_2((x+1)(x-1))=3.

Rewrite in exponential form:

(x+1)(x-1)=2^3=8

, so

x^2-1=8

, giving

x^2=9

, hence

x=\pm3

. Now apply the domain:

x+1>0

and

x-1>0

, so

x>-1

and

x>1

. These together mean

x>1

. Thus

-3

is extraneous, and the only valid solution is

3

. Why the others fail:

\pm3

includes the extraneous negative root, and

-3

alone is invalid because it makes both

x+1=-2

and

x-1=-4

, neither of which is a valid log input. The choice

9

likely comes from seeing

x^2=9

and jumping to

x=9

; remember that solving

x^2=9

gives

x=\pm3

, not

9

. Study tip: whenever a log equation leads to multiple candidate solutions, always plug them back into the original expressions or check the domain before finalizing. On these exams, extraneous roots from squared equations are a classic trap.

Question 7

If $p=\log_4 7$ , which expression is equivalent to $\log_2 49$ ?

$\frac{p}{2}$
$2p$
$4p$ (correct answer)
$8p$

Explanation: When you see logarithms with different bases, your first thought should be to rewrite them using the same base. Here, notice that

49=7^2

and

4=2^2

, so both

7

and

49

are powers of

7

, while the bases are powers of

2

. Start with

\log_2 49 = \log_2(7^2) = 2\log_2 7

. Now relate

\log_2 7

p=\log_4 7

. Using the change-of-base formula,

\log_4 7 = \frac{\log_2 7}{\log_2 4} = \frac{\log_2 7}{2}

. So

\log_2 7 = 2p

. Therefore,

\log_2 49 = 2(2p)=4p

. The choice

\frac{p}{2}

would come from dividing by 2 instead of multiplying, essentially computing

\log_4 \sqrt{7}

. The choice

2p

equals

\log_4 49

, not

\log_2 49

: it changes the number but leaves the base as 4. The choice

8p

would be

8\log_4 7 = \log_4(7^8)

, far too large; it may come from multiplying both the exponent 2 and the base conversion factor 2 incorrectly. A strong strategy: whenever a log has a base that is a power of another base, convert using

\log_{a^k} x = \frac{1}{k}\log_a x

. Here,

\log_4 7 = \frac{1}{2}\log_2 7

, so solving for

\log_2 7

gives

2p

. Then apply the exponent rule

\log(x^n)=n\log x

. Recognizing these two log properties quickly turns this into a two-step problem.

Question 8

Suppose $\log_b 2 = p$ , $\log_b 3 = q$ , and $\log_b 5 = r$ . Which expression is equivalent to $\log_b\left(\frac{5\sqrt{5}}{12}\right)$ ?

$\frac{3}{2}r - 2p - q$ (correct answer)
$\frac{3}{2}r - p - q$
$\frac{5}{2}r - 2p - q$
$\frac{3}{2}r - 2p - \frac{1}{2}q$

Explanation: Whenever you see a logarithm of a fraction involving products and roots, your first move is to expand using the properties:

\log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N

and

\log_b(M^k)=k\log_b M

. Here, rewrite the expression carefully. Notice that

5\sqrt{5} = 5^{3/2}

because

\sqrt{5}=5^{1/2}

, and

12 = 3 \cdot 4 = 3 \cdot 2^2

. So the log becomes:

\log_b\left(\frac{5^{3/2}}{3\cdot 2^2}\right) = \frac{3}{2}\log_b 5 - \log_b 3 - 2\log_b 2 = \frac{3}{2}r - q - 2p.

That matches the choice

\frac{3}{2}r - 2p - q

, which is correct. Now, why are the others off? The choice

\frac{3}{2}r - p - q

loses the factor of 2 in front of

p

— it treats

12

3\cdot 2

instead of

3\cdot 2^2

. The choice

\frac{5}{2}r - 2p - q

comes from thinking

5\sqrt{5} = 5^{5/2}

, but that would be

5^2\cdot 5^{1/2}

, not

5^1\cdot 5^{1/2}

. Finally,

\frac{3}{2}r - 2p - \frac{1}{2}q

incorrectly halves the

q

term — the denominator has

3

, not

\sqrt{3}

. Your takeaway: before expanding a logarithm, rewrite any radicals as fractional exponents and factor out all integers completely (like

12 = 2^2 \cdot 3

). Then apply the power rule to each term, paying special attention to coefficients. This avoids the most common traps on this exam.

Question 9

Which value is equivalent to $\log_2 3 \cdot \log_3 5 \cdot \log_5 8$ ?

$1$
$2$
$3$ (correct answer)
$8$

Explanation: When you see a product of logarithms with different bases, think about the change-of-base formula:

\log_a b = \frac{\ln b}{\ln a}

. Write each log in terms of natural logs:

\frac{\ln 3}{\ln 2}\cdot \frac{\ln 5}{\ln 3}\cdot \frac{\ln 8}{\ln 5}

The

\ln 3

and

\ln 5

cancel, leaving

\frac{\ln 8}{\ln 2} = \log_2 8

. Since

2^3 = 8

, this equals

3

. So the correct value is

3

. The choice

1

is what you'd get if you mistakenly thought all the logs somehow "cancel out" to leave

1

. The choice

2

is a trap if you focus on the bases

2,3,5

and think there's some averaging, but there is no such operation. The choice

8

is tempting if you confuse the numeric result with the last argument in the expression; the log product is an exponent-like quantity not equal to the final base's power. In fact,

\log_2 8

gives

3

, not

8

. A strong strategy for logarithm products like this is to rewrite every logarithm using a common base and look for telescoping cancellation. If you see a chain

\log_a b \cdot \log_b c \cdot \log_c d

, the answer is always

\log_a d

. This saves time and avoids errors on exam problems that test the change-of-base formula.

Question 10

Which expression is equivalent to $\ln\left(\frac{e^2\sqrt{x}}{(x+1)^3}\right)$ for $x>0$ ?

$2+\frac{1}{2}\ln x - 3\ln(x+1)$ (correct answer)
$2+\frac{1}{2}\ln x + 3\ln(x+1)$
$\frac{1}{2}\ln x - 3\ln(x+1) - 2$
$2+\ln x - 3\ln(x+1)$

Explanation: Whenever you see a logarithm of a fraction containing products, powers, and roots, apply the three core log rules:

\ln(ab)=\ln a+\ln b

\ln(a/b)=\ln a-\ln b

, and

\ln(a^r)=r\ln a

. Here, break the expression apart before simplifying. Start with the numerator:

\ln(e^2\sqrt{x})=\ln(e^2)+\ln(\sqrt{x})

. Since

\ln(e^2)=2

and

\sqrt{x}=x^{1/2}

, this becomes

2+\frac12\ln x

. Then subtract the log of the denominator:

\ln((x+1)^3)=3\ln(x+1)

. Because the denominator is divided, you subtract it, giving

2+\frac12\ln x-3\ln(x+1).

This matches the expression

2+\frac{1}{2}\ln x - 3\ln(x+1)

exactly. Now look at the wrong choices. The expression

2+\frac{1}{2}\ln x + 3\ln(x+1)

uses a plus sign before the denominator term, but division inside the logarithm becomes subtraction, not addition. The expression

\frac{1}{2}\ln x - 3\ln(x+1) - 2

uses

-2

instead of

+2

; the

e^2

is in the numerator, so its log is positive

2

. Finally,

2+\ln x - 3\ln(x+1)

treats

\sqrt{x}

x

, forgetting that the square root means the exponent is

\frac12

, not

1

. A useful habit: rewrite roots as fractional exponents first, then expand the logarithm term by term. Check every sign carefully — the most common trap on this exam is losing the subtraction for a denominator or misreading a root as a plain power.

Question 11

If $u=\log_3 x$ , which expression is equivalent to $\log_x 27$ ?

$\frac{u}{3}$
$\frac{3}{u}$ (correct answer)
$3u$
$u^3$

Explanation: Whenever you see a logarithm with a variable base, your first move should be to rewrite everything in exponential form or use the change-of-base idea. Since

u=\log_3 x

, you know

x=3^u

. Also,

27=3^3

, so the expression becomes

\log_x 27=\log_{3^u}(3^3)=\frac{\ln(3^3)}{\ln(3^u)}=\frac{3\ln 3}{u\ln 3}=\frac{3}{u}.

So the correct expression is

\frac{3}{u}

. Now look at the traps. The choice

\frac{u}{3}

is actually

\log_{27}x

, not

\log_x 27

; it's the reciprocal of the correct answer, often produced when you confuse which quantity is the base. The choice

3u

comes from noticing

27=3^3

and multiplying by

\log_3 x

, but that would represent

\log_3(x^3)

, not a log with base

x

. Finally,

u^3

treats the logarithm like a number to be cubed, which confuses the exponent on the input with the log value itself. A strong strategy for this exam: when you see logs with bases that are powers of one another, rewrite all bases as a common base and compare exponents. Here,

x=3^u

and

27=3^3

, so the answer is the ratio of those exponents:

\frac{3}{u}

Question 12

Given $\log_{12} 2 = a$ and $\log_{12} 3 = b$ , which expression is equivalent to $\log_6 8$ ?

$\frac{3a}{1-b}$
$\frac{3a}{a-b}$
$\frac{3a}{1-a-b}$
$\frac{3a}{a+b}$ (correct answer)

Explanation: Whenever you see a logarithm in an unfamiliar base, your first move is to change to a base you know. Here all given logs are base

12

, so rewrite

\log_6 8

using base

12

\log_6 8=\frac{\log_{12} 8}{\log_{12} 6}

Now simplify each piece. Since

8=2^3

\log_{12} 8=\log_{12}(2^3)=3\log_{12}2=3a

And since

6=2\cdot 3

\log_{12} 6=\log_{12}2+\log_{12}3=a+b

So the expression is

\frac{3a}{a+b}

The key was recognizing

6=2\cdot 3

, making the denominator a sum. Now look at the traps. The choice

\frac{3a}{1-b}

uses

1-b=\log_{12}12-\log_{12}3=\log_{12}4

, not

\log_{12}6

. The choice

\frac{3a}{a-b}

uses

a-b=\log_{12}(2/3)

, again not

\log_{12}6

. The choice

\frac{3a}{1-a-b}

simplifies

1-a-b=\log_{12}(12/(2\cdot3))=\log_{12}2=a

, so it becomes

3

, not

\log_6 8

. Each wrong option comes from mixing up product and quotient log rules. Study tip: after changing base, simplify the numerator and denominator separately. Remember

\log(xy)=\log x+\log y

and

\log(x/y)=\log x-\log y

. The denominator must equal the original base's input exactly — here,

6