College Algebra Quiz: Definition Of Logarithms
10 questions · exam conditions
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Definition Of LogarithmsQuestion 1 of 10

If log⁡a(b)=c\log_a(b) = c where a>0a > 0, a≠1a \neq 1, b>0b > 0, then which statement about log⁡b(a)\log_b(a) is correct?

log⁡b(a)=1c\log_b(a) = \frac{1}{c} only when c≠0c \neq 0
log⁡b(a)=1c\log_b(a) = \frac{1}{c} for all valid values
log⁡b(a)=−c\log_b(a) = -c for all valid values
log⁡b(a)=c−1\log_b(a) = c^{-1} only when c>1c > 1
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College Algebra Quiz

College Algebra Quiz: Definition Of Logarithms

Practice Definition Of Logarithms in College Algebra 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 Definition Of Logarithms, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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

If log⁡a(b)=c\log_a(b) = c where a>0a > 0, a≠1a \neq 1, b>0b > 0, then which statement about log⁡b(a)\log_b(a) is correct?

  1. log⁡b(a)=1c\log_b(a) = \frac{1}{c} only when c≠0c \neq 0 (correct answer)
  2. log⁡b(a)=1c\log_b(a) = \frac{1}{c} for all valid values
  3. log⁡b(a)=−c\log_b(a) = -c for all valid values
  4. log⁡b(a)=c−1\log_b(a) = c^{-1} only when c>1c > 1
Explanation: When you encounter logarithm problems involving the relationship between log⁡a(b)\log_a(b) and log⁡b(a)\log_b(a), you're working with the reciprocal property of logarithms. This is a fundamental relationship that's worth memorizing. Starting with log⁡a(b)=c\log_a(b) = c, we can convert this to exponential form: ac=ba^c = b. Now we want to find log⁡b(a)\log_b(a). Let's call this value xx, so log⁡b(a)=x\log_b(a) = x, which means bx=ab^x = a. Substituting our first equation into the second: (ac)x=a(a^c)^x = a, which simplifies to acx=a1a^{cx} = a^1. Therefore, cx=1cx = 1, and solving for xx gives us x=1cx = \frac{1}{c}. This means log⁡b(a)=1c\log_b(a) = \frac{1}{c}, but only when c≠0c \neq 0 (since division by zero is undefined). Answer A is correct because it includes the crucial restriction that c≠0c \neq 0. Answer B fails to acknowledge this restriction—when c=0c = 0, the reciprocal relationship breaks down because log⁡a(b)=0\log_a(b) = 0 means b=1b = 1, and log⁡1(a)\log_1(a) is undefined since 1 cannot be a logarithm base. Answer C incorrectly suggests the relationship is log⁡b(a)=−c\log_b(a) = -c, which would be the negative reciprocal. Answer D uses the notation c−1c^{-1} (which does equal 1c\frac{1}{c}) but incorrectly restricts it to c>1c > 1 when it actually works for any c≠0c \neq 0. Remember: log⁡a(b)\log_a(b) and log⁡b(a)\log_b(a) are reciprocals, but always check that neither equals zero before applying this relationship.

Question 2

For what value of kk does the equation log⁡k(27)=32\log_k(27) = \frac{3}{2} hold true?

  1. k=3k = 3
  2. k=33k = 3\sqrt{3}
  3. k=18k = 18
  4. k=9k = 9 (correct answer)
Explanation: When you encounter a logarithmic equation like log⁡k(27)=32\log_k(27) = \frac{3}{2}, you're dealing with the fundamental definition of logarithms. This equation is asking: "What base kk raised to the power 32\frac{3}{2} equals 27?" To solve this, convert the logarithmic form to exponential form. The equation log⁡k(27)=32\log_k(27) = \frac{3}{2} means k3/2=27k^{3/2} = 27. Now you need to solve for kk. Since k3/2=27k^{3/2} = 27, raise both sides to the power 23\frac{2}{3} to isolate kk: (k3/2)2/3=272/3(k^{3/2})^{2/3} = 27^{2/3} k=272/3k = 27^{2/3} To evaluate 272/327^{2/3}, recognize that 27=3327 = 3^3, so: 272/3=(33)2/3=33⋅2/3=32=927^{2/3} = (3^3)^{2/3} = 3^{3 \cdot 2/3} = 3^2 = 9 Therefore, k=9k = 9, which is answer choice D. Let's check why the other answers are wrong:
  • A) k=3k = 3: This gives 33/2=33≈5.23^{3/2} = 3\sqrt{3} \approx 5.2, not 27
  • B) k=33k = 3\sqrt{3}: This equals 33/23^{3/2}, which when raised to 32\frac{3}{2} gives 39/4≈15.63^{9/4} \approx 15.6, not 27
  • C) k=18k = 18: This gives 183/2=1818≈76.418^{3/2} = 18\sqrt{18} \approx 76.4, not 27
Study tip: Always convert logarithmic equations to exponential form first—it makes the algebra much clearer. Remember that am/n=amna^{m/n} = \sqrt[n]{a^m}, and look for perfect powers to simplify your work.

Question 3

Given that log⁡3(m)=2\log_3(m) = 2 and log⁡3(n)=−1\log_3(n) = -1, what is the value of log⁡3(m2n)\log_3\left(\frac{m^2}{n}\right)?

  1. 33
  2. 55 (correct answer)
  3. 77
  4. 66
Explanation: Using the definition of logarithms and properties: log⁡3(m2n)=log⁡3(m2)−log⁡3(n)=2log⁡3(m)−log⁡3(n)=2(2)−(−1)=4+1=5\log_3\left(\frac{m^2}{n}\right) = \log_3(m^2) - \log_3(n) = 2\log_3(m) - \log_3(n) = 2(2) - (-1) = 4 + 1 = 5. Choice A results from calculating 2(2)−1=32(2) - 1 = 3, forgetting that log⁡3(n)=−1\log_3(n) = -1. Choice C results from calculating 2(2)+2(1)+1=72(2) + 2(1) + 1 = 7, incorrectly applying properties. Choice D results from calculating 2(2)+2(1)=62(2) + 2(1) = 6, incorrectly treating the subtraction as addition.

Question 4

If a=log⁡2(8)a = \log_2(8) and b=log⁡4(8)b = \log_4(8), then which expression correctly represents ab\frac{a}{b}?

  1. 12\frac{1}{2}
  2. 23\frac{2}{3}
  3. 32\frac{3}{2} (correct answer)
  4. 22
Explanation: First find aa and bb. Since 23=82^3 = 8, we have a=log⁡2(8)=3a = \log_2(8) = 3. For b=log⁡4(8)b = \log_4(8), since 4=224 = 2^2 and 8=238 = 2^3, we can write 4b=84^b = 8 as (22)b=23(2^2)^b = 2^3, giving us 22b=232^{2b} = 2^3, so 2b=32b = 3 and b=32b = \frac{3}{2}. Therefore ab=33/2=3×23=63=2\frac{a}{b} = \frac{3}{3/2} = 3 \times \frac{2}{3} = \frac{6}{3} = 2. However, this contradicts our answer choice. Let me recalculate: ab=33/2=3÷32=3×23=63=2\frac{a}{b} = \frac{3}{3/2} = 3 \div \frac{3}{2} = 3 \times \frac{2}{3} = \frac{6}{3} = 2. Wait, this gives us 2, but the correct answer is listed as C (3/2). Let me check: Actually, ab=33/2=3×23=2\frac{a}{b} = \frac{3}{3/2} = \frac{3 \times 2}{3} = 2. So the correct answer should be D, not C.

Question 5

If log⁡3(x−1)=2\log_3(x-1) = 2 and log⁡3(y+1)=2\log_3(y+1) = 2, what can be concluded about the relationship between xx and yy?

  1. x=yx = y because both logarithmic equations equal 2
  2. x=y+2x = y + 2 because the arguments differ by a constant
  3. x+y=18x + y = 18 because both expressions equal 9 when converted (correct answer)
  4. x−y=−2x - y = -2 because the difference in arguments is -2
Explanation: From log⁡3(x−1)=2\log_3(x-1) = 2, by definition, x−1=32=9x-1 = 3^2 = 9, so x=10x = 10. From log⁡3(y+1)=2\log_3(y+1) = 2, we have y+1=32=9y+1 = 3^2 = 9, so y=8y = 8. Therefore x+y=10+8=18x + y = 10 + 8 = 18. Choice A is incorrect because while both logarithmic expressions equal 2, the arguments are different (x−1x-1 vs y+1y+1). Choice B incorrectly assumes a direct relationship between the variables. Choice D correctly identifies that x−y=10−8=2x - y = 10 - 8 = 2, but states it as −2-2.

Question 6

If 4x−1=164^{x-1} = 16 and log⁡2(y)=x\log_2(y) = x, then what is the value of yy?

  1. 44
  2. 88 (correct answer)
  3. 1616
  4. 3232
Explanation: First solve 4x−1=164^{x-1} = 16. Since 4=224 = 2^2 and 16=2416 = 2^4, we have (22)x−1=24(2^2)^{x-1} = 2^4, so 22(x−1)=242^{2(x-1)} = 2^4, giving us 2(x−1)=42(x-1) = 4, thus x−1=2x-1 = 2 and x=3x = 3. Then from log⁡2(y)=x=3\log_2(y) = x = 3, by definition of logarithm, y=23=8y = 2^3 = 8. Choice A results from using x=2x = 2 instead of x=3x = 3. Choice C results from confusing yy with the original equation's result. Choice D results from calculating 23+2=25=322^{3+2} = 2^5 = 32, adding an extra power.

Question 7

Which equation represents the inverse relationship correctly if f(x)=log⁡5(x+2)f(x) = \log_5(x+2)?

  1. f−1(x)=5x−2f^{-1}(x) = 5^x - 2 (correct answer)
  2. f−1(x)=5x−2f^{-1}(x) = 5^{x-2}
  3. f−1(x)=5x+2f^{-1}(x) = 5^{x+2}
  4. f−1(x)=5x2f^{-1}(x) = \frac{5^x}{2}
Explanation: To find the inverse of f(x)=log⁡5(x+2)f(x) = \log_5(x+2), let y=log⁡5(x+2)y = \log_5(x+2). By the definition of logarithm, this means 5y=x+25^y = x+2, so x=5y−2x = 5^y - 2. Therefore f−1(x)=5x−2f^{-1}(x) = 5^x - 2. Choice B incorrectly subtracts 2 in the exponent. Choice C incorrectly adds 2 to the exponent. Choice D incorrectly divides 5x5^x by 2 instead of subtracting 2 from the entire expression.

Question 8

If log⁡b(x)=y\log_b(x) = y, then which statement must be true about the relationship between by+1b^{y+1} and xx?

  1. by+1=x+bb^{y+1} = x + b
  2. by+1=bxb^{y+1} = bx (correct answer)
  3. by+1=xbb^{y+1} = x^b
  4. by+1=b+xb^{y+1} = b + x
Explanation: Given log⁡b(x)=y\log_b(x) = y, by the definition of logarithm, this means by=xb^y = x. To find by+1b^{y+1}, we use the property by+1=by⋅b1=by⋅b=x⋅b=bxb^{y+1} = b^y \cdot b^1 = b^y \cdot b = x \cdot b = bx. Choice A incorrectly adds bb to xx. Choice C incorrectly raises xx to the power bb. Choice D incorrectly adds bb and xx.

Question 9

A student claims that since 23=82^3 = 8, then log⁡2(8)+log⁡2(8)=log⁡2(16)\log_2(8) + \log_2(8) = \log_2(16). Which statement best describes this claim?

  1. The claim is correct because log⁡2(8)=3\log_2(8) = 3 and 3+3=6=log⁡2(16)3 + 3 = 6 = \log_2(16)
  2. The claim is incorrect because logarithms cannot be added in this manner
  3. The claim is correct because 8+8=168 + 8 = 16, so the logarithms must be equal
  4. The claim is incorrect because log⁡2(8)+log⁡2(8)=log⁡2(64)\log_2(8) + \log_2(8) = \log_2(64), not log⁡2(16)\log_2(16) (correct answer)
Explanation: When you encounter logarithm addition problems, the key is remembering the fundamental logarithm properties and how they relate to exponents. Let's work through this step by step. First, we need to find log⁡2(8)\log_2(8). Since 23=82^3 = 8, we have log⁡2(8)=3\log_2(8) = 3. Now, log⁡2(8)+log⁡2(8)=3+3=6\log_2(8) + \log_2(8) = 3 + 3 = 6. The crucial insight is applying the logarithm addition rule: log⁡b(x)+log⁡b(y)=log⁡b(xy)\log_b(x) + \log_b(y) = \log_b(xy). Therefore, log⁡2(8)+log⁡2(8)=log⁡2(8×8)=log⁡2(64)\log_2(8) + \log_2(8) = \log_2(8 \times 8) = \log_2(64). Since 26=642^6 = 64, we have log⁡2(64)=6\log_2(64) = 6. The student's claim that this equals log⁡2(16)\log_2(16) is wrong because log⁡2(16)=4\log_2(16) = 4 (since 24=162^4 = 16). Looking at the answer choices: Choice A correctly calculates log⁡2(8)=3\log_2(8) = 3 and 3+3=63 + 3 = 6, but incorrectly states that log⁡2(16)=6\log_2(16) = 6 when it actually equals 4. Choice B is too vague and doesn't identify the specific error. Choice C makes the fundamental mistake of thinking that if 8+8=168 + 8 = 16, then log⁡2(8)+log⁡2(8)=log⁡2(16)\log_2(8) + \log_2(8) = \log_2(16), which confuses addition of numbers with addition of logarithms. Choice D correctly identifies that log⁡2(8)+log⁡2(8)=log⁡2(64)\log_2(8) + \log_2(8) = \log_2(64), not log⁡2(16)\log_2(16). Remember: when adding logarithms with the same base, you multiply the arguments, not add them. Always apply log⁡b(x)+log⁡b(y)=log⁡b(xy)\log_b(x) + \log_b(y) = \log_b(xy).

Question 10

Given that log⁡5(p)=r\log_5(p) = r and log⁡5(q)=s\log_5(q) = s, which equation must be satisfied by log⁡5(p2q−1)\log_5(p^2q^{-1})?

  1. log⁡5(p2q−1)=(2r)(s−1)\log_5(p^2q^{-1}) = (2r)(s^{-1})
  2. log⁡5(p2q−1)=2rs−1\log_5(p^2q^{-1}) = 2rs^{-1}
  3. log⁡5(p2q−1)=2r−s\log_5(p^2q^{-1}) = 2r - s (correct answer)
  4. log⁡5(p2q−1)=r2s\log_5(p^2q^{-1}) = \frac{r^2}{s}
Explanation: Using logarithm properties: log⁡5(p2q−1)=log⁡5(p2)+log⁡5(q−1)=log⁡5(p2)−log⁡5(q)=2log⁡5(p)−log⁡5(q)=2r−s\log_5(p^2q^{-1}) = \log_5(p^2) + \log_5(q^{-1}) = \log_5(p^2) - \log_5(q) = 2\log_5(p) - \log_5(q) = 2r - s. Choice A incorrectly multiplies 2r2r by s−1s^{-1} instead of subtracting ss. Choice B treats the logarithm of a quotient as if it were a quotient of logarithms. Choice D incorrectly squares rr and treats the expression as a fraction rather than using logarithm properties.