Math 3 Quiz: Logarithm Properties

19 questions · exam conditions

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Logarithm PropertiesQuestion 1 of 19

Which expression is equivalent to $2\log_6(x) - \log_6(9) + \log_6(3)$ when $x > 0$ ?

\log_6(\frac{x^2}{27})

\log_6(\frac{x^2 \cdot 3}{9})

\log_6(\frac{x^2}{3})

\log_6(x^2) - 2

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Math 3 Quiz

Math 3 Quiz: Logarithm Properties

Practice Logarithm Properties in Math 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 Math 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 $2\log_6(x) - \log_6(9) + \log_6(3)$ when $x > 0$ ?

$\log_6(\frac{x^2}{27})$
$\log_6(\frac{x^2 \cdot 3}{9})$
$\log_6(\frac{x^2}{3})$ (correct answer)
$\log_6(x^2) - 2$

Explanation: When you encounter logarithmic expressions with multiple terms, your goal is to use logarithm properties to combine them into a single logarithm. The key properties you need are:

a\log_b(x) = \log_b(x^a)

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

, and

\log_b(x) - \log_b(y) = \log_b(\frac{x}{y})

. Let's work through

2\log_6(x) - \log_6(9) + \log_6(3)

step by step. First, apply the power property:

2\log_6(x) = \log_6(x^2)

. This gives us

\log_6(x^2) - \log_6(9) + \log_6(3)

. Next, combine the addition and subtraction using the product and quotient properties. Since we have subtraction followed by addition, we get:

\log_6\left(\frac{x^2 \cdot 3}{9}\right)

. Simplifying the fraction:

\frac{x^2 \cdot 3}{9} = \frac{3x^2}{9} = \frac{x^2}{3}

. Therefore, the answer is

\log_6\left(\frac{x^2}{3}\right)

, which is choice C. Choice A gives

\frac{x^2}{27}

, which would result from incorrectly treating

+\log_6(3)

-\log_6(3)

. Choice B shows

\frac{x^2 \cdot 3}{9}

without simplifying the fraction—this is the intermediate step, not the final answer. Choice D attempts to convert

\log_6(9)

incorrectly, since

\log_6(9) \neq 2

. Remember: always simplify fractions completely when working with logarithms, and double-check that you're applying addition and subtraction properties correctly—the order matters!

Question 2

If $\log_2(x+3) - \log_2(x-1) = 2$ , then $x$ equals:

$\frac{11}{3}$
$5$
$7$
$\frac{7}{3}$ (correct answer)

Explanation: Using the quotient property:

\log_2\left(\frac{x+3}{x-1}\right) = 2

. This means

\frac{x+3}{x-1} = 2^2 = 4

. Cross-multiplying:

x + 3 = 4(x-1) = 4x - 4

. Solving:

x + 3 = 4x - 4

, so

7 = 3x

, giving

x = \frac{7}{3}

. We must check that

x > 1

for the domain:

\frac{7}{3} \approx 2.33 > 1

✓. Choice A, B, and C result from arithmetic errors in solving the linear equation.

Question 3

Which expression is equivalent to $\log_3\left(\frac{9x^2}{\sqrt{y}}\right)$ when $x > 0$ and $y > 0$ ?

$2 + 2\log_3(x) - \frac{1}{2}\log_3(y)$ (correct answer)
$2 + \log_3(x^2) - \log_3(\sqrt{y})$
$\log_3(9) + \log_3(x^2) - \log_3(y^{1/2})$
$2\log_3(3) + 2\log_3(x) + \frac{1}{2}\log_3(y)$

Explanation: Using logarithm properties:

\log_3\left(\frac{9x^2}{\sqrt{y}}\right) = \log_3(9) + \log_3(x^2) - \log_3(\sqrt{y}) = \log_3(3^2) + 2\log_3(x) - \log_3(y^{1/2}) = 2 + 2\log_3(x) - \frac{1}{2}\log_3(y)

. Choice B doesn't fully expand the power rule, choice C doesn't simplify

\log_3(9) = 2

, and choice D has an incorrect sign.

Question 4

Given that $\log_5(2) \approx 0.431$ , which value is closest to $\log_5(0.8)$ ?

$0.334$
$0.097$
$-0.334$
$-0.097$ (correct answer)

Explanation: When you encounter logarithm problems with decimal inputs, the key is recognizing how to break down the number using logarithm properties, especially when you're given a related logarithm value. To find

\log_5(0.8)

, start by rewriting 0.8 as a fraction:

0.8 = \frac{8}{10} = \frac{4}{5}

. Now you can use the logarithm quotient rule:

\log_5(0.8) = \log_5\left(\frac{4}{5}\right) = \log_5(4) - \log_5(5)

. Since

\log_5(5) = 1

, you have

\log_5(0.8) = \log_5(4) - 1

. Next, recognize that

4 = 2^2

, so

\log_5(4) = \log_5(2^2) = 2\log_5(2)

. Using the given value:

\log_5(4) = 2 \times 0.431 = 0.862

. Therefore:

\log_5(0.8) = 0.862 - 1 = -0.138

, which is closest to

-0.097

. Option A (0.334) represents a common error of forgetting the negative sign and miscalculating the final arithmetic. Option B (0.097) makes the same arithmetic mistake as the correct answer but incorrectly assumes the result should be positive. Option C (-0.334) gets the negative sign right but likely confused the calculation, perhaps by incorrectly using

\log_5(2) - 1

directly. Remember that logarithms of numbers between 0 and 1 are always negative, and practice breaking down decimal numbers into fractions to apply logarithm properties effectively.

Question 5

If $\log_2(x) + \log_2(y) = 5$ and $\log_2(x) - \log_2(y) = 1$ , what is the value of $xy$ ?

16
32 (correct answer)
64
128

Explanation: Using logarithm properties: From the given equations,

\log_2(xy) = \log_2(x) + \log_2(y) = 5

, so

xy = 2^5 = 32

. We can verify by solving the system: adding the equations gives

2\log_2(x) = 6

, so

\log_2(x) = 3

and

x = 8

. Substituting back:

\log_2(y) = 2

, so

y = 4

. Thus

xy = 32

. Choice A uses

2^4

instead of

2^5

. Choice C incorrectly calculates

x + y = 12

then squares it. Choice D uses

2^7

by incorrectly adding both original equations.

Question 6

If $\log_5(m) = p$ and $\log_5(n) = q$ , which expression represents $\log_5(\frac{25m^3}{n^2})$ ?

$2 + 3p - 2q$ (correct answer)
$25 + 3p - 2q$
$\frac{25 + 3p}{2q}$
$2 + \frac{3p}{2q}$

Explanation: Using logarithm properties:

\log_5(\frac{25m^3}{n^2}) = \log_5(25) + \log_5(m^3) - \log_5(n^2) = \log_5(5^2) + 3\log_5(m) - 2\log_5(n) = 2 + 3p - 2q

. Choice B incorrectly treats

\log_5(25)

as 25 instead of 2. Choice C misapplies the quotient rule as algebraic division. Choice D incorrectly applies the quotient rule to the variables

p

and

q

Question 7

If $\log_2(x) + \log_2(y) = 5$ and $\log_2(x) - \log_2(y) = 1$ , what is the value of $xy$ ?

32 (correct answer)
16
8
4

Explanation: Using logarithm properties:

\log_2(x) + \log_2(y) = \log_2(xy) = 5

, so

xy = 2^5 = 32

. We can verify by solving the system: adding the equations gives

2\log_2(x) = 6

, so

\log_2(x) = 3

and

x = 8

. Substituting back:

\log_2(8) + \log_2(y) = 5

, so

3 + \log_2(y) = 5

, giving

y = 4

. Therefore

xy = 32

Question 8

Which statement best explains why $\log_a(xy) = \log_a(x) + \log_a(y)$ requires both $x > 0$ and $y > 0$ ?

Negative numbers would make the logarithm base invalid for this property
The product $xy$ must be positive for the equation to hold algebraically
Logarithms are only defined for positive real numbers in their standard form (correct answer)
The addition of logarithms only works when both arguments are positive integers

Explanation: When working with logarithmic properties, you need to understand the fundamental domain restrictions that make these functions well-defined. The logarithm function

\log_a(x)

exists only when its argument is a positive real number—this is a foundational requirement, not just a convenient restriction. The property

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

requires both

x > 0

and

y > 0

because logarithms themselves are only defined for positive real numbers in their standard form. If either

x

y

were negative or zero, then

\log_a(x)

\log_a(y)

would be undefined, making the entire equation meaningless. This makes answer C correct. Let's examine why the other options miss the mark. Answer A incorrectly focuses on the base

a

, but the base isn't what's problematic here—it's the arguments

x

and

y

that must be positive. Answer B suggests the issue is purely algebraic about the product

xy

, but even if

xy > 0

(say, when both

x

and

y

are negative), you still couldn't evaluate

\log_a(x)

\log_a(y)

individually. Answer D incorrectly restricts the domain to positive integers, when logarithms work for all positive real numbers. Remember this key principle: before applying any logarithm property, check that every logarithm in the equation is actually defined. The domain of logarithms—positive real numbers only—often determines what values are permissible in logarithmic equations and properties.

Question 9

Which equation demonstrates the correct application of the change of base formula to rewrite $\log_8(64)$ ?

$\log_8(64) = \log(64) - \log(8)$
$\log_8(64) = \frac{\log(8)}{\log(64)}$
$\log_8(64) = \frac{\log(64)}{\log(8)}$ (correct answer)
$\log_8(64) = \frac{\log_8(64)}{\log_8(10)}$

Explanation: When you encounter logarithms with bases other than 10 or e, the change of base formula becomes essential for solving problems with a calculator or converting between different forms. The change of base formula states that

\log_b(a) = \frac{\log(a)}{\log(b)}

, where the logarithms on the right can be in any convenient base (typically base 10). To apply this to

\log_8(64)

, you place 64 (the argument) in the numerator and 8 (the original base) in the denominator:

\log_8(64) = \frac{\log(64)}{\log(8)}

. This is exactly what option C shows. Option A incorrectly applies the quotient rule for logarithms, which would give you

\log\left(\frac{64}{8}\right) = \log(8)

, not

\log_8(64)

. The quotient rule only works when both logarithms have the same base. Option B flips the fraction, putting the original base in the numerator and the argument in the denominator. This reversal would actually give you

\frac{1}{\log_{64}(8)}

, which is completely different from what we want. Option D attempts to use the change of base formula but keeps the same base (8) for both the original expression and the denominator. This creates a circular definition that doesn't actually change the base at all. Remember the change of base formula as "new over old" - the original argument goes over the original base. This pattern will help you avoid the common trap of reversing the fraction, which appears frequently on math exams.

Question 10

Given $\log_3(a) = 2$ and $\log_3(b) = -1$ , what is $\log_3\left(\frac{a^2}{b^3}\right)$ ?

$-1$
$1$
$7$ (correct answer)
$5$

Explanation: When you encounter logarithmic expressions with given logarithm values, you'll need to apply the fundamental properties of logarithms to break down complex expressions into simpler parts. Start by using the logarithm properties:

\log_b(x^n) = n\log_b(x)

and

\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)

. This means

\log_3\left(\frac{a^2}{b^3}\right) = \log_3(a^2) - \log_3(b^3) = 2\log_3(a) - 3\log_3(b)

. Now substitute the given values:

\log_3(a) = 2

and

\log_3(b) = -1

. This gives you

2\log_3(a) - 3\log_3(b) = 2(2) - 3(-1) = 4 - (-3) = 4 + 3 = 7

. Looking at the wrong answers: Choice A (

-1

) likely comes from incorrectly calculating

2(-1) - 3(2) = -5

, mixing up the given values. Choice B (

1

) might result from errors like

2(2) - 3(1) = 1

, possibly confusing

\log_3(b) = -1

with

\log_3(b) = 1

. Choice D (

5

) could come from sign errors, such as calculating

2(2) - 3(-1) = 4 - 3 = 1

incorrectly, or

2(2) + 3(-1) = 4 - 3 = 1

then adding instead of subtracting. The answer is C (

7

). Study tip: Always write out the logarithm properties step-by-step before substituting values. This prevents sign errors and helps you track which operations to perform when dealing with quotients and exponents in logarithmic expressions.

Question 11

If $\log_b(x) = 3\log_b(2) + \frac{1}{2}\log_b(9) - 2\log_b(3)$ , then $x$ equals:

$3$
$\frac{24}{9}$
$8$
$\frac{8}{3}$ (correct answer)

Explanation: When you encounter logarithmic expressions with multiple terms, the key is using logarithm properties to simplify before solving. The three essential properties are:

n\log_b(a) = \log_b(a^n)

\log_b(a) + \log_b(c) = \log_b(ac)

, and

\log_b(a) - \log_b(c) = \log_b(\frac{a}{c})

. Let's apply these properties step by step. First, simplify each term using the power rule:

3\log_b(2) = \log_b(2^3) = \log_b(8)

\frac{1}{2}\log_b(9) = \log_b(9^{1/2}) = \log_b(3)

, and

2\log_b(3) = \log_b(3^2) = \log_b(9)

. Now the equation becomes:

\log_b(x) = \log_b(8) + \log_b(3) - \log_b(9)

Using the addition and subtraction properties:

\log_b(x) = \log_b(\frac{8 \cdot 3}{9}) = \log_b(\frac{24}{9}) = \log_b(\frac{8}{3})

Since the logarithms are equal,

x = \frac{8}{3}

, which is choice D. Choice A (

3

) likely comes from incorrectly simplifying

\log_b(9)^{1/2}

. Choice B (

\frac{24}{9}

) results from forgetting to reduce the fraction or misapplying the subtraction property. Choice C (

8

) occurs when students ignore the other terms and focus only on

3\log_b(2) = \log_b(8)

. Remember: always convert coefficients to exponents first, then combine logarithms using addition/subtraction rules. Work systematically through each property rather than trying shortcuts.

Question 12

If $\log_a(m) = p$ and $\log_a(n) = q$ , which expression represents $\log_a\left(\frac{m^3}{n^2\sqrt[4]{a}}\right)$ ?

$3p - 2q - \frac{1}{4}$ (correct answer)
$3p - 2q + \frac{1}{4}$
$\frac{3p}{2q} - \frac{1}{4}$
$3p - 2q - 4$

Explanation: Using logarithm properties:

\log_a\left(\frac{m^3}{n^2\sqrt[4]{a}}\right) = \log_a(m^3) - \log_a(n^2) - \log_a(a^{1/4}) = 3\log_a(m) - 2\log_a(n) - \frac{1}{4}\log_a(a) = 3p - 2q - \frac{1}{4}

. Choice B has wrong sign, choice C incorrectly uses division instead of subtraction, choice D uses

-4

instead of

-\frac{1}{4}

Question 13

If $\log_6(18) + \log_6(2) = \log_6(k)$ , then $k$ equals:

$20$
$36$ (correct answer)
$108$
$216$

Explanation: Using the product property:

\log_6(18) + \log_6(2) = \log_6(18 \cdot 2) = \log_6(36)

. Therefore

k = 36

. Choice A comes from

18 + 2 = 20

(incorrectly adding arguments), choice C from

18 \cdot 6 = 108

(multiplying by base instead of the other argument), choice D from

6^3 = 216

(misremembering that

6^2 = 36

Question 14

Which statement correctly justifies why $\log_2(a-b) = \log_2(a) - \log_2(b)$ is generally false?

The logarithm of a difference equals the quotient of logarithms, not their difference
The logarithm of a difference cannot be simplified using standard logarithm properties alone (correct answer)
The logarithm properties only apply when the arguments are positive integers
The logarithm of a difference equals the difference of logarithms only when $a > b$

Explanation: The logarithm of a difference

\log_2(a-b)

cannot be simplified using the standard logarithm properties (product, quotient, power rules). These properties apply to products, quotients, and powers, not sums or differences of the arguments. Choice A confuses this with quotient property, choice C incorrectly limits logarithm properties to positive integers, choice D incorrectly suggests the false equation works under certain conditions.

Question 15

For which value of $k$ is the equation $\log_2(x + 1) + \log_2(x - 1) = \log_2(k)$ equivalent to $x^2 - 1 = k$ ?

All positive values of $k$
All values of $k$ such that $k > 0$ and $k \neq 1$
All values of $k$ such that $k \geq 0$
All values of $k$ such that $k > 0$ and $x > 1$ (correct answer)

Explanation: Using the product rule:

\log_2(x + 1) + \log_2(x - 1) = \log_2((x + 1)(x - 1)) = \log_2(x^2 - 1)

. So the equation becomes

\log_2(x^2 - 1) = \log_2(k)

, which gives

x^2 - 1 = k

. However, for the original equation to be valid, we need

x + 1 > 0

and

x - 1 > 0

, so

x > 1

. This restricts the domain and thus the possible values of

k

. Choice A ignores domain restrictions. Choice B considers

k > 0

but misses the

x > 1

constraint. Choice C incorrectly includes

k = 0

Question 16

Given that $\log_3(x) = 2$ and $\log_9(y) = \frac{1}{2}$ , what is the value of $\log_3(\frac{xy}{3})$ ?

$\frac{3}{2}$
$\frac{5}{2}$
$2$ (correct answer)
$\frac{7}{2}$

Explanation: First, convert

\log_9(y) = \frac{1}{2}

to base 3: Since

9 = 3^2

, we have

\log_9(y) = \frac{\log_3(y)}{\log_3(9)} = \frac{\log_3(y)}{2} = \frac{1}{2}

, so

\log_3(y) = 1

. Now:

\log_3(\frac{xy}{3}) = \log_3(x) + \log_3(y) - \log_3(3) = 2 + 1 - 1 = 2

. Choice A forgets to subtract

\log_3(3) = 1

. Choice B adds all terms without subtracting. Choice D incorrectly computes

\log_3(y) = \frac{3}{2}

Question 17

Which statement correctly justifies why $\log_3(x - 2) + \log_3(x + 2) = \log_3(x^2 - 4)$ is not valid for all real numbers $x$ ?

The equation requires $x^2 - 4 > 0$ , so $x < -2$ or $x > 2$
The equation requires $x > 0$ for the logarithm to be defined
The equation requires $x \neq \pm 2$ to avoid division by zero
The equation requires $x - 2 > 0$ and $x + 2 > 0$ , so $x > 2$ (correct answer)

Explanation: When working with logarithmic equations, you must always consider the domain restrictions before applying algebraic properties. Logarithms are only defined for positive arguments, so each logarithmic term imposes its own constraint. For this equation to be valid, both

\log_3(x - 2)

and

\log_3(x + 2)

must be defined. This requires:

$x - 2 > 0$ , which means $x > 2$
$x + 2 > 0$ , which means $x > -2$

Since both conditions must be satisfied simultaneously, you need

x > 2

. Only when

x > 2

can you apply the logarithm property

\log_a(m) + \log_a(n) = \log_a(mn)

to get

\log_3(x^2 - 4)

. This makes choice D correct. Choice A identifies the correct inequality

x^2 - 4 > 0

(which gives

x < -2

x > 2

), but this only ensures the right side is defined. The left side has stricter requirements that eliminate

x < -2

. Choice B suggests

x > 0

is sufficient, but this misses the crucial point that

x - 2

must be positive, not just

x

itself. Choice C mentions avoiding division by zero, but there's no division in this logarithmic equation. The issue is about keeping arguments positive, not avoiding zero denominators. Remember: when analyzing logarithmic equations, always check the domain of each individual logarithmic term first. The intersection of all domain restrictions determines where the equation is valid, regardless of what algebraic manipulations might suggest.

Question 18

Which expression is equivalent to $\log_3(\frac{27x^2}{\sqrt{y}})$ when $x > 0$ and $y > 0$ ?

$3 + 2\log_3(x) - \frac{1}{2}\log_3(y)$ (correct answer)
$3 + 2\log_3(x) - 2\log_3(y)$
$\frac{27 + 2\log_3(x)}{\frac{1}{2}\log_3(y)}$
$27 + \log_3(x^2) - \log_3(\sqrt{y})$

Explanation: Using logarithm properties:

\log_3(\frac{27x^2}{\sqrt{y}}) = \log_3(27) + \log_3(x^2) - \log_3(\sqrt{y}) = \log_3(3^3) + 2\log_3(x) - \log_3(y^{1/2}) = 3 + 2\log_3(x) - \frac{1}{2}\log_3(y)

. Choice B incorrectly applies the power rule to

\sqrt{y}

2\log_3(y)

. Choice C misapplies the quotient rule as division of expressions. Choice D fails to evaluate

\log_3(27)

and doesn't apply power rules.

Question 19

If $\log_a(b) = 3$ and $\log_a(c) = -2$ , what is the value of $\log_a(\frac{b^2\sqrt{c}}{a})$ ?

$5$
$4$ (correct answer)
$3$
$2$

Explanation: Using logarithm properties:

\log_a(\frac{b^2\sqrt{c}}{a}) = \log_a(b^2) + \log_a(\sqrt{c}) - \log_a(a) = 2\log_a(b) + \frac{1}{2}\log_a(c) - 1 = 2(3) + \frac{1}{2}(-2) - 1 = 6 - 1 - 1 = 4

. Choice A forgets to subtract

\log_a(a) = 1

. Choice C incorrectly calculates

2(3) + \frac{1}{2}(-2) = 6 - 1 = 5

then subtracts 2 instead of 1. Choice D makes an error in the power rule for

\sqrt{c}