Algebra 2 Quiz: Using Logarithm Properties
20 questions · exam conditions
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Using Logarithm PropertiesQuestion 1 of 20

Use the logarithm properties log⁡b(xy)=log⁡b(x)+log⁡b(y),log⁡b(xy)=log⁡b(x)−log⁡b(y),log⁡b(xp)=plog⁡b(x)\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x) to evaluate without a calculator: log⁡3(819).\log_3\left(\frac{81}{9}\right).

12\tfrac{1}{2}
00
22
44
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Algebra 2 Quiz

Algebra 2 Quiz: Using Logarithm Properties

Practice Using Logarithm Properties in Algebra 2 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 Using Logarithm Properties, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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

Use the logarithm properties log⁡b(xy)=log⁡b(x)+log⁡b(y),log⁡b(xy)=log⁡b(x)−log⁡b(y),log⁡b(xp)=plog⁡b(x)\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x) to evaluate without a calculator: log⁡3(819).\log_3\left(\frac{81}{9}\right).

  1. 12\tfrac{1}{2}
  2. 00
  3. 22 (correct answer)
  4. 44
Explanation: This question tests your understanding of the quotient property of logarithms and your ability to evaluate logs by recognizing powers of the base. The quotient property states that log_b(x/y) = log_b(x) - log_b(y), which means you can split a fraction inside a log into a difference, or evaluate it directly if you recognize the result as a power of the base. To evaluate log_3(81/9): First, calculate the fraction: 81/9 = 9. Then evaluate log_3(9): Since 9 = 3², we have log_3(9) = log_3(3²) = 2. Alternatively, using the quotient property: log_3(81/9) = log_3(81) - log_3(9) = log_3(3⁴) - log_3(3²) = 4 - 2 = 2. Choice C correctly gives 2 as the answer, whether you simplify first or use the quotient property. Choice A (1/2) would mean 3^(1/2) = 81/9, but 3^(1/2) = √3 ≈ 1.73, not 9. The key insight is recognizing powers of 3: 3¹ = 3, 3² = 9, 3³ = 27, 3⁴ = 81. Evaluation strategy for quotients: (1) You can simplify the fraction first if possible (81/9 = 9), then evaluate the simpler log, (2) Or apply quotient property and evaluate each log separately, (3) Both methods give the same answer—choose based on which recognition is easier. Mental math tip: memorize small powers of common bases (2, 3, 5, 10) to quickly evaluate many logarithms without a calculator!

Question 2

Simplify using logarithm properties: log⁡4(64)−log⁡4(4)\log_4(64) - \log_4(4)

  1. 11
  2. 22 (correct answer)
  3. 33
  4. 44
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—to simplify differences of logs into a single value. The three logarithm properties come from exponent rules and work because logarithms are exponents: (1) Product property: log⁡b(xy)=log⁡b(x)+log⁡b(y)\log_b(xy) = \log_b(x) + \log_b(y), (2) Quotient property: log⁡b(x)−log⁡b(y)=log⁡b(x/y)\log_b(x) - \log_b(y) = \log_b(x/y), (3) Power property: log⁡b(xp)=plog⁡b(x)\log_b(x^p) = p \log_b(x). To simplify log⁡4(64)−log⁡4(4)\log_4(64) - \log_4(4), apply quotient: log⁡4(64/4)=log⁡4(16)\log_4(64/4) = \log_4(16); since 16=4216 = 4^2, log⁡4(16)=2\log_4(16) = 2. Choice B correctly uses the quotient property and evaluates by rewriting as a power of the base. Choice A might come from subtracting the arguments directly—remember to combine logs first! Simplify by combining with properties before evaluating, and express arguments as powers of the base when possible. This makes numerical answers pop out easily—excellent effort!

Question 3

Write the expression as a single logarithm (condense using log properties):

2log⁡10(a)+12log⁡10(b)−log⁡10(c)2\log_{10}(a)+\tfrac{1}{2}\log_{10}(b)-\log_{10}(c)

(Assume a>0a>0, b>0b>0, and c>0c>0.)

  1. log⁡10(a2bc)\log_{10}\left(\frac{a^2\sqrt{b}}{c}\right) (correct answer)
  2. log⁡10(2a+bc)\log_{10}\left(\frac{2a+\sqrt{b}}{c}\right)
  3. log⁡10(a2bc)\log_{10}\left(\frac{a^2}{bc}\right)
  4. log⁡10(a b2c)\log_{10}\left(\frac{a\,b^{2}}{c}\right)
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: log_b(xy) = log_b(x) + log_b(y) because multiplying numbers means adding their exponents (b^m × b^n = b^(m+n)), (2) Quotient property: log_b(x/y) = log_b(x) - log_b(y) because dividing means subtracting exponents (b^m ÷ b^n = b^(m-n)), (3) Power property: log_b(xpx^p) = p·log_b(x) because raising to power means multiplying exponents ((b^m)^p = b^(mp)). These properties transform multiplication into addition, division into subtraction, and exponentiation into multiplication—powerful simplification tools! To condense 2log₁₀(a) + ½log₁₀(b) - log₁₀(c), we reverse the properties: (1) Apply power property backwards: 2log₁₀(a) becomes log₁₀(a²) because coefficient 2 goes up as exponent, and ½log₁₀(b) becomes log₁₀(b^(1/2)) = log₁₀(√b). (2) Now we have log₁₀(a²) + log₁₀(√b) - log₁₀(c). (3) Apply product property backwards to the addition: log₁₀(a²) + log₁₀(√b) = log₁₀(a²√b). (4) Apply quotient property backwards to the subtraction: log₁₀(a²√b) - log₁₀(c) = log₁₀(a²√b/c). Final condensed form: log₁₀(a²√b/c)! Choice A correctly applies the logarithm properties in reverse order to condense into a single logarithm. Choice B incorrectly interprets the coefficients as regular multiplication: 2log(a) does NOT become log(2a), it becomes log(a²). The coefficient becomes an exponent, not a multiplier inside the log! Condensing strategy (reverse process): (1) Use power property backwards: coefficient in front becomes exponent (3log(x) becomes log(x³)). (2) Identify additions: log(a) + log(b) condenses to log(ab) using product property backwards. (3) Identify subtractions: log(a) - log(b) condenses to log(a/b) using quotient property backwards. (4) Combine all: the result should be single log of one big expression. Example: 2log(x) + log(y) - 3log(z) becomes log(x²) + log(y) - log(z³), then log(x²y) - log(z³), finally log(x²y/z³). Work step-by-step combining two logs at a time!

Question 4

Use logarithm properties to evaluate without a calculator:

log⁡3(819)\log_3\left(\frac{81}{9}\right)​

  1. 00
  2. 11
  3. 22 (correct answer)
  4. 33
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The quotient property states log_b(x/y) = log_b(x) - log_b(y), and when combined with recognizing powers of the base, you can evaluate logarithms without a calculator! To evaluate log_3(81/9): Method 1 using quotient property: (1) Apply quotient property: log_3(81/9) = log_3(81) - log_3(9), (2) Recognize 81 = 3⁴ and 9 = 3², (3) So log_3(81) = 4 and log_3(9) = 2, (4) Therefore: 4 - 2 = 2. Method 2 by simplifying first: (1) Calculate 81/9 = 9, (2) So log_3(81/9) = log_3(9), (3) Since 9 = 3², we have log_3(9) = 2. Choice C correctly gives the answer 2. Choice A (0) would only be correct if 81/9 = 1, but 81/9 = 9, not 1—remember log_b(1) = 0 for any base b, but we don't have log_3(1) here! Evaluation strategy for logarithms with quotients: (1) You can either apply quotient property first, then evaluate each log separately, or (2) Simplify the fraction first, then take the logarithm of the result, (3) Express all numbers as powers of the base when possible (81 = 3⁴, 9 = 3², 27 = 3³), (4) Use log_b(bnb^n) = n to get your answer. Both methods give the same result, so choose whichever seems easier for the specific problem!

Question 5

Use the logarithm properties log⁡b(xy)=log⁡b(x)+log⁡b(y),log⁡b(xy)=log⁡b(x)−log⁡b(y),log⁡b(xp)=plog⁡b(x)\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x) to expand the expression as a sum and/or difference of logarithms: log⁡3(x2yz3).\log_3\left(\frac{x^2y}{z^3}\right).

  1. log⁡3(x2)+log⁡3(y)+log⁡3(z3)\log_3(x^2)+\log_3(y)+\log_3(z^3)
  2. 2log⁡3(x)+log⁡3(y)−3log⁡3(z)2\log_3(x)+\log_3(y)-3\log_3(z) (correct answer)
  3. log⁡3(x)+log⁡3(y)−log⁡3(z)\log_3(x)+\log_3(y)-\log_3(z)
  4. log⁡3(x2+y)−log⁡3(z3)\log_3(x^2+y)-\log_3(z^3)
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences. The three logarithm properties come from exponent rules and work because logarithms are exponents: (1) Product property: log_b(xy) = log_b(x) + log_b(y) because multiplying numbers means adding their exponents, (2) Quotient property: log_b(x/y) = log_b(x) - log_b(y) because dividing means subtracting exponents, (3) Power property: log_b(xpx^p) = p log_b(x) because raising to power means multiplying exponents. To expand log_3((x2x^2 y)/z^3), first apply the quotient property: log_3(x2x^2 y) - log_3(z3z^3), then apply the product property to the numerator: log_3(x2x^2) + log_3(y) - log_3(z3z^3), and finally the power property: 2 log_3(x) + log_3(y) - 3 log_3(z). Choice B correctly applies these properties in sequence to fully expand the expression into a sum and difference of simpler logs with coefficients. Choice D incorrectly combines x^2 + y inside a log instead of using the product property for x^2 y, remember there's no property for log of a sum—it's a common mix-up with the product rule! When expanding logarithms, start with the outermost operation like quotients or overall powers, then break down products and individual powers step by step. This systematic approach ensures you don't miss any terms—keep practicing, you've got this!

Question 6

Use the logarithm properties log⁡b(xy)=log⁡b(x)+log⁡b(y)\log_b(xy)=\log_b(x)+\log_b(y), log⁡b(xy)=log⁡b(x)−log⁡b(y)\log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y), and log⁡b(xp)=plog⁡b(x)\log_b(x^p)=p\log_b(x) to expand the expression: log⁡5(x3yz2).\log_5\left(\frac{x^3\sqrt{y}}{z^2}\right). Write your answer as a sum/difference of logarithms.

  1. 3log⁡5(x)+12log⁡5(y)−2log⁡5(z)3\log_5(x)+\frac{1}{2}\log_5(y)-2\log_5(z) (correct answer)
  2. log⁡5(x3)+log⁡5(y)+log⁡5(z2)\log_5(x^3)+\log_5(\sqrt{y})+\log_5(z^2)
  3. log⁡5(x3)+log⁡5(y)−log⁡5(z)\log_5(x^3)+\log_5(\sqrt{y})-\log_5(z)
  4. log⁡5(x+yz)\log_5\left(\frac{x+y}{z}\right)
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: log⁡b(xy)=log⁡b(x)+log⁡b(y)\log_b(xy) = \log_b(x) + \log_b(y) because multiplying numbers means adding their exponents, (2) Quotient property: log⁡b(x/y)=log⁡b(x)−log⁡b(y)\log_b(x/y) = \log_b(x) - \log_b(y) because dividing means subtracting exponents, (3) Power property: log⁡b(xp)=plog⁡b(x)\log_b(x^p) = p \log_b(x) because raising to power means multiplying exponents. These properties transform multiplication into addition, division into subtraction, and exponentiation into multiplication—powerful simplification tools! To expand log⁡5((x3y)/z2)\log_5((x^3 \sqrt{y})/z^2) using properties: (1) Apply quotient property first: log⁡5(x3y)−log⁡5(z2)\log_5(x^3 \sqrt{y}) - \log_5(z^2) because the overall structure is a fraction. (2) Apply product property to first log: log⁡5(x3)+log⁡5(y)\log_5(x^3) + \log_5(\sqrt{y}) because x3yx^3 \sqrt{y} is a product. (3) Apply power property to each: 3log⁡5(x)+12log⁡5(y)−2log⁡5(z)3\log_5(x) + \frac{1}{2}\log_5(y) - 2\log_5(z) because x3x^3 has exponent 3, y\sqrt{y} is y1/2y^{1/2}, and z2z^2 has exponent 2. Final expanded form: 3log⁡5(x)+12log⁡5(y)−2log⁡5(z)3\log_5(x) + \frac{1}{2}\log_5(y) - 2\log_5(z). Choice A correctly applies the logarithm properties in the right order to expand into sum and difference of simpler logs. Choice B incorrectly keeps the logs condensed without fully expanding using the power property, leaving exponents inside instead of bringing them out as coefficients—remember to apply power rule to complete the expansion! Expanding logarithm strategy: (1) Identify overall structure: is it product, quotient, or power? Start with outermost operation. (2) Apply matching property: multiplication becomes +, division becomes -, exponent becomes coefficient. (3) Work from outside in: handle fraction first (quotient property), then products within (product property), then exponents (power property). (4) Continue until each log has single variable with coefficient. You're doing great—keep practicing this step-by-step approach to master log expansions!

Question 7

Use the logarithm properties log⁡b(xy)=log⁡b(x)+log⁡b(y),log⁡b(xy)=log⁡b(x)−log⁡b(y),log⁡b(xp)=plog⁡b(x)\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x) to expand the expression log⁡((mn)4p)\log\left(\frac{(mn)^4}{p}\right) (common log base 10) into a sum/difference of logs with no exponents inside the logs.

  1. 4log⁡(mn)−log⁡(p)4\log(mn)-\log(p)
  2. 4log⁡(m)+4log⁡(n)−log⁡(p)4\log(m)+4\log(n)-\log(p) (correct answer)
  3. log⁡(m)+log⁡(n)−log⁡(p4)\log(m)+\log(n)-\log(p^4)
  4. log⁡(m+n)4−log⁡(p)\log(m+n)^4-\log(p)
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences with no exponents inside the logs. The three properties are: (1) Product property: log(xy) = log(x) + log(y), (2) Quotient property: log(x/y) = log(x) - log(y), (3) Power property: log(xpx^p) = p·log(x)—this one is crucial for removing exponents from inside logs. To expand log((mn)⁴/p): First apply the quotient property to the overall fraction: log((mn)⁴) - log(p). Next apply the power property to the first term: 4log(mn) - log(p), bringing the exponent 4 down as a coefficient. Then apply the product property inside: 4[log(m) + log(n)] - log(p). Finally distribute the 4: 4log(m) + 4log(n) - log(p). Choice B correctly shows the fully expanded form with no exponents inside any logarithm and the coefficient 4 distributed to both m and n. Choice A incorrectly leaves mn together inside the log instead of separating them—when you have 4log(mn), you must apply the product property to get 4log(m) + 4log(n), not just 4log(mn). The key insight is that (mn)⁴ = m⁴n⁴, so log((mn)⁴) = log(m⁴n⁴) = log(m⁴) + log(n⁴) = 4log(m) + 4log(n). Expansion strategy for removing all exponents: (1) Handle the overall structure first (quotient here), (2) Bring down any exponents as coefficients using power property, (3) Expand any products or quotients inside, (4) Distribute coefficients if needed. The goal is to have each log contain only a single variable with no exponent, with any powers appearing as coefficients in front!

Question 8

Use the logarithm properties log⁡b(xy)=log⁡b(x)+log⁡b(y),log⁡b(xy)=log⁡b(x)−log⁡b(y),log⁡b(xp)=plog⁡b(x)\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x) to write as a single logarithm: log⁡2(5)+log⁡2(3)−log⁡2(4).\log_2(5)+\log_2(3)-\log_2(4).

  1. log⁡2(154)\log_2\left(\dfrac{15}{4}\right) (correct answer)
  2. log⁡2(512)\log_2\left(\dfrac{5}{12}\right)
  3. log⁡2(5+3−4)\log_2(5+3-4)
  4. log⁡2(158)\log_2\left(\dfrac{15}{8}\right)
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you condense sums and differences of logarithms into a single logarithm. The properties for condensing are: (1) Product property backwards: log_b(x) + log_b(y) = log_b(xy), so addition becomes multiplication, (2) Quotient property backwards: log_b(x) - log_b(y) = log_b(x/y), so subtraction becomes division. To condense log_2(5) + log_2(3) - log_2(4): First combine the addition: log_2(5) + log_2(3) = log_2(5·3) = log_2(15). Then handle the subtraction: log_2(15) - log_2(4) = log_2(15/4). Choice A correctly shows the condensed form as log_2(15/4), a single logarithm of the fraction 15/4. Choice C incorrectly tries to combine using 5 + 3 - 4 inside the log, but there is NO logarithm property that allows log(a) + log(b) - log(c) = log(a + b - c). You must use multiplication for addition of logs and division for subtraction of logs! Condensing strategy: (1) Work left to right combining two logs at a time, (2) Addition of logs → multiply the arguments, (3) Subtraction of logs → divide the arguments, (4) Never add or subtract the arguments directly. The pattern to remember: plus outside means times inside, minus outside means divide inside!

Question 9

Using the properties log⁡b(xy)=log⁡b(x)+log⁡b(y),log⁡b(xy)=log⁡b(x)−log⁡b(y),log⁡b(xp)=plog⁡b(x),\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x), write the expression as a single logarithm (condense): 2log⁡5(a)−12log⁡5(b)+log⁡5(c).2\log_5(a)-\tfrac{1}{2}\log_5(b)+\log_5(c).

  1. log⁡5(a2cb2)\log_5\left(\frac{a^2c}{b^2}\right)
  2. log⁡5(a2)+log⁡5(c)−log⁡5(b)\log_5(a^2)+\log_5(c)-\log_5(\sqrt{b})
  3. log⁡5(a2cb)\log_5\left(\frac{a^2c}{\sqrt{b}}\right) (correct answer)
  4. log⁡5(a2bc)\log_5\left(\frac{a^2\sqrt{b}}{c}\right)
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: logb(xy)=logb(x)+logb(y)log_b(xy) = log_b(x) + log_b(y) because multiplying numbers means adding their exponents, (2) Quotient property: logb(x/y)=logb(x)−logb(y)log_b(x/y) = log_b(x) - log_b(y) because dividing means subtracting exponents, (3) Power property: logb(xp)=p⋅logb(x)log_b(x^p) = p\cdot log_b(x) because raising to power means multiplying exponents. To condense 2log5(a)−(1/2)log5(b)+log5(c)2log_5(a) - (1/2)log_5(b) + log_5(c), we work backwards: (1) Apply power property backwards: 2log5(a)2log_5(a) becomes log5(a2)log_5(a^2) and (1/2)log5(b)(1/2)log_5(b) becomes log5(b(1/2))=log5(b)log_5(b^(1/2)) = log_5(\sqrt{b}), (2) Rewrite: log5(a2)−log5(b)+log5(c)log_5(a^2) - log_5(\sqrt{b}) + log_5(c), (3) Apply quotient property backwards to first two: log5(a2/b)+log5(c)log_5(a^2/\sqrt{b}) + log_5(c), (4) Apply product property backwards: log5((a2/b)⋅c)=log5(a2c/b)log_5((a^2/\sqrt{b})\cdot c) = log_5(a^2c/\sqrt{b}). Choice A correctly condenses the expression into a single logarithm by applying properties in reverse order. Choice B incorrectly has b\sqrt{b} in the numerator instead of denominator—remember that subtraction of logs means division, so −log5(b)-log_5(\sqrt{b}) puts b\sqrt{b} in the denominator! Condensing strategy (reverse process): (1) Use power property backwards: coefficient in front becomes exponent (2log(x)2log(x) becomes log(x2)log(x^2)), (2) Identify subtractions: log(a)−log(b)log(a) - log(b) condenses to log(a/b)log(a/b) using quotient property backwards, (3) Identify additions: log(a)+log(b)log(a) + log(b) condenses to log(ab)log(ab) using product property backwards, (4) Combine all: the result should be single log of one big expression. Work step-by-step combining two logs at a time, and be careful with signs—minus means divide!

Question 10

Express using the change-of-base formula (in terms of natural logs):

log⁡5(20)\log_5(20)

  1. ln⁡(5)ln⁡(20)\dfrac{\ln(5)}{\ln(20)}
  2. ln⁡(20)ln⁡(5)\dfrac{\ln(20)}{\ln(5)} (correct answer)
  3. ln⁡(20)−ln⁡(5)\ln(20)-\ln(5)
  4. ln⁡(100)\ln(100)
Explanation: This question tests your understanding of the change-of-base formula, which allows you to express logarithms with any base in terms of logarithms with a more convenient base (like base 10 or base e). The change-of-base formula states: log_b(x) = log_c(x)/log_c(b) for any valid base c. This works because if b^y = x, then taking log_c of both sides gives y·log_c(b) = log_c(x), so y = log_c(x)/log_c(b). Most commonly, we use natural logarithm (ln) or common logarithm (log₁₀) as the new base because calculators have these functions built in! To express log₅(20) using natural logs: Apply the change-of-base formula with c = e: log₅(20) = ln(20)/ln(5). This converts the base-5 logarithm into a ratio of natural logarithms that can be evaluated on any scientific calculator. The formula essentially asks: "If 5^y = 20, what is y?" and solves it using natural logs. Choice B correctly applies the change-of-base formula as ln(20)/ln(5). Choice A incorrectly inverts the fraction—it shows ln(5)/ln(20), but the correct formula has the argument (20) in the numerator and the base (5) in the denominator. Remember: log_base(argument) = ln(argument)/ln(base). The argument goes on top! Change-of-base strategy: (1) Identify what you're converting: log_b(x) where b is the base and x is the argument. (2) Choose your new base (usually e for ln or 10 for log). (3) Apply formula: log_b(x) = ln(x)/ln(b) or log_b(x) = log(x)/log(b). (4) The argument x goes in numerator, base b goes in denominator—don't mix them up! This formula is essential for calculating logarithms with unusual bases on a calculator that only has ln and log₁₀ buttons.

Question 11

Use the logarithm properties log⁡b(xy)=log⁡b(x)+log⁡b(y),log⁡b(xy)=log⁡b(x)−log⁡b(y),log⁡b(xp)=plog⁡b(x)\log_b(xy)=\log_b(x)+\log_b(y),\quad \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\quad \log_b(x^p)=p\log_b(x) to simplify the expression: log⁡5(125x)log⁡5(5).\frac{\log_5(125x)}{\log_5(5)}.

  1. log⁡5(125x)\log_5(125x)
  2. 3+log⁡5(x)3+\log_5(x) (correct answer)
  3. log⁡5(3x)\log_5(3x)
  4. 3log⁡5(x)\dfrac{3}{\log_5(x)}
Explanation: This question tests your understanding of logarithm properties and the special case when the denominator is log_b(b) = 1, which allows for elegant simplification. The key insight is that log_5(5) = 1 because 5¹ = 5, so the expression becomes log_5(125x)/1 = log_5(125x). To simplify log_5(125x)/log_5(5): First recognize that log_5(5) = 1, so we have log_5(125x)/1 = log_5(125x). Next, apply the product property: log_5(125x) = log_5(125) + log_5(x). Then evaluate log_5(125): Since 125 = 5³, we have log_5(125) = log_5(5³) = 3. Therefore: log_5(125x) = 3 + log_5(x). Choice B correctly shows the simplified form as 3 + log_5(x), recognizing that 125 = 5³ contributes a value of 3. Choice A just writes log_5(125x) without simplifying—while not wrong, it misses the opportunity to evaluate log_5(125) = 3. Choice C incorrectly suggests log_5(125) = log_5(3), but 125 = 5³, not 3·5. Simplification strategy when log_b(b) appears: (1) Always remember log_b(b) = 1 for any base b, (2) This often creates opportunities to simplify further, (3) Look for perfect powers of the base to evaluate (like 125 = 5³ here), (4) Express the final answer in simplest form by evaluating any logs of perfect powers. This type of problem rewards recognizing special values and powers of the base!

Question 12

Use the change-of-base formula (optional) and log properties to rewrite log⁡4(25)\log_4(25) in terms of natural logarithms. You may use log⁡b(x)=ln⁡(x)ln⁡(b).\log_b(x)=\frac{\ln(x)}{\ln(b)}.

  1. ln⁡(4)ln⁡(25)\dfrac{\ln(4)}{\ln(25)}
  2. ln⁡(25)ln⁡(4)\dfrac{\ln(25)}{\ln(4)} (correct answer)
  3. ln⁡(100)\ln(100)
  4. ln⁡(25)−ln⁡(4)\ln(25)-\ln(4)
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The change-of-base formula states that log_b(x) = ln(x)/ln(b), which converts any logarithm to natural logarithms (or any other base). To rewrite log_4(25) in terms of natural logarithms: Apply the change-of-base formula directly: log_4(25) = ln(25)/ln(4). Choice B correctly applies the change-of-base formula with numerator ln(25) and denominator ln(4). Choice A incorrectly flips the fraction—remember that the argument (25) goes in the numerator and the base (4) goes in the denominator of the change-of-base formula! The change-of-base formula is incredibly useful: (1) It lets you evaluate any logarithm using a calculator that only has ln or log_10, (2) The pattern is always log_b(x) = ln(x)/ln(b)—argument on top, base on bottom, (3) You can use any new base, not just e: log_b(x) = log_c(x)/log_c(b) for any valid base c. This formula connects all logarithms regardless of base, showing they're all related by simple division!

Question 13

Which statement about log⁡4(x3y2z1/2)\log_4\left(\frac{x^3y^2}{z^{1/2}}\right) is correct when x,y,z>0x, y, z > 0?

  1. It equals 3log⁡4(x)+2log⁡4(y)12log⁡4(z)\frac{3\log_4(x) + 2\log_4(y)}{\frac{1}{2}\log_4(z)}
  2. It equals 3log⁡4(x)+2log⁡4(y)+12log⁡4(z)3\log_4(x) + 2\log_4(y) + \frac{1}{2}\log_4(z)
  3. It cannot be simplified using logarithm properties
  4. It equals 3log⁡4(x)+2log⁡4(y)−12log⁡4(z)3\log_4(x) + 2\log_4(y) - \frac{1}{2}\log_4(z) (correct answer)
Explanation: When you encounter a logarithm with a complex argument involving multiplication, division, and exponents, you should immediately think about applying the three fundamental logarithm properties to break it down into simpler parts. Let's work through log⁡4(x3y2z1/2)\log_4\left(\frac{x^3y^2}{z^{1/2}}\right) step by step. First, use the quotient rule: log⁡b(MN)=log⁡b(M)−log⁡b(N)\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N). This gives us: log⁡4(x3y2)−log⁡4(z1/2)\log_4(x^3y^2) - \log_4(z^{1/2}) Next, apply the product rule to the first term: log⁡b(MN)=log⁡b(M)+log⁡b(N)\log_b(MN) = \log_b(M) + \log_b(N): log⁡4(x3)+log⁡4(y2)−log⁡4(z1/2)\log_4(x^3) + \log_4(y^2) - \log_4(z^{1/2}) Finally, use the power rule log⁡b(Mn)=nlog⁡b(M)\log_b(M^n) = n\log_b(M) on each term: 3log⁡4(x)+2log⁡4(y)−12log⁡4(z)3\log_4(x) + 2\log_4(y) - \frac{1}{2}\log_4(z) This confirms answer D is correct. Choice A incorrectly treats the division as creating a fraction with the logarithms themselves, which violates logarithm properties. Choice B makes a sign error by adding 12log⁡4(z)\frac{1}{2}\log_4(z) instead of subtracting it—this happens when you forget that logarithm of a quotient involves subtraction. Choice C is wrong because logarithm expressions can almost always be simplified using the fundamental properties. Study tip: Remember the acronym "PQP" for logarithm properties: Product (addition), Quotient (subtraction), Power (multiplication by exponent). Always apply quotient rule first when you see division, and watch your signs carefully—division in the argument becomes subtraction between logarithms.

Question 14

Expand the expression using logarithm properties:

log⁡(10xyz2)\log\left(\frac{10x\sqrt{y}}{z^2}\right)

(Here log⁡\log is base 10.)​

  1. log⁡(10)+log⁡(x)+12log⁡(y)−2log⁡(z)\log(10)+\log(x)+\tfrac{1}{2}\log(y)-2\log(z) (correct answer)
  2. log⁡(10x)+log⁡(y−z2)\log(10x)+\log(\sqrt{y}-z^2)
  3. log⁡(10)+log⁡(x)+log⁡(y)−log⁡(z2)\log(10)+\log(x)+\log(y)-\log(z^2)
  4. log⁡(10)+log⁡(x)−12log⁡(y)−2log⁡(z)\log(10)+\log(x)-\tfrac{1}{2}\log(y)-2\log(z)
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules: (1) Product property: log_b(xy) = log_b(x) + log_b(y), (2) Quotient property: log_b(x/y) = log_b(x) - log_b(y), (3) Power property: log_b(xpx^p) = p·log_b(x), and remember that √y = y^(1/2), so log(√y) = (1/2)log(y)! To expand log(10x√y/z²): (1) Apply quotient property first: log(10x√y) - log(z²) because the overall structure is a fraction, (2) Apply product property to numerator: log(10) + log(x) + log(√y) because 10x√y is a product of three factors, (3) Apply power property: log(10) + log(x) + (1/2)log(y) - 2log(z) because √y = y^(1/2) and z² has exponent 2. Choice A correctly applies all properties to get log(10) + log(x) + (1/2)log(y) - 2log(z). Choice B incorrectly tries to combine √y - z² inside a single logarithm, but this makes no sense—you can't have log(√y - z²) from a quotient! The quotient property gives subtraction of logarithms, not subtraction inside a logarithm. Expanding logarithm strategy with radicals: (1) Convert radicals to fractional exponents first: √y = y^(1/2), ∛x = x^(1/3), etc., (2) Apply quotient property to handle overall fraction structure, (3) Apply product property to break up multiplications, (4) Apply power property last, bringing all exponents (including fractions) down as coefficients. This systematic approach handles even complex expressions with roots and powers!

Question 15

Condense to a single logarithm:

log⁡5(x)−2log⁡5(y)+log⁡5(z)\log_5(x)-2\log_5(y)+\log_5(z)

(Assume x>0x>0, y>0y>0, and z>0z>0.)

  1. log⁡5(xzy2)\log_5\left(\frac{xz}{y^2}\right) (correct answer)
  2. log⁡5(x2yz)\log_5\left(\frac{x}{2yz}\right)
  3. log⁡5(xy2z)\log_5\left(\frac{x}{y^2 z}\right)
  4. log⁡5(xy2z)\log_5\left(\frac{xy^2}{z}\right)
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: log_b(xy) = log_b(x) + log_b(y) because multiplying numbers means adding their exponents (b^m × b^n = b^(m+n)), (2) Quotient property: log_b(x/y) = log_b(x) - log_b(y) because dividing means subtracting exponents (b^m ÷ b^n = b^(m-n)), (3) Power property: log_b(xpx^p) = p·log_b(x) because raising to power means multiplying exponents ((b^m)^p = b^(mp)). These properties transform multiplication into addition, division into subtraction, and exponentiation into multiplication—powerful simplification tools! To condense log₅(x) - 2log₅(y) + log₅(z), we reverse the properties: (1) Apply power property backwards to -2log₅(y): this becomes -log₅(y²) because the coefficient 2 goes up as an exponent. (2) Now we have log₅(x) - log₅(y²) + log₅(z). (3) Rearrange as log₅(x) + log₅(z) - log₅(y²) to group additions first. (4) Apply product property backwards to addition: log₅(x) + log₅(z) = log₅(xz). (5) Apply quotient property backwards: log₅(xz) - log₅(y²) = log₅(xz/y²). Final condensed form: log₅(xz/y²)! Choice A correctly applies the logarithm properties in reverse order to condense into a single logarithm. Choice C incorrectly handles the coefficient: -2log₅(y) becomes -log₅(y²), not -log₅(y)·2. The coefficient becomes an exponent on y, giving y² in the denominator, not y²z in the denominator. Watch how coefficients become exponents! Condensing strategy (reverse process): (1) Use power property backwards: coefficient in front becomes exponent (3log(x) becomes log(x³)). (2) Identify additions: log(a) + log(b) condenses to log(ab) using product property backwards. (3) Identify subtractions: log(a) - log(b) condenses to log(a/b) using quotient property backwards. (4) Combine all: the result should be single log of one big expression. Remember that negative coefficients still become exponents: -2log(y) = -log(y²), then the minus sign indicates it goes in the denominator!

Question 16

Simplify using logarithm properties:

log⁡7(49x)−log⁡7(7)\log_7(49x)-\log_7(7)

(Assume x>0x>0.)

  1. log⁡7(7x)\log_7(7x) (correct answer)
  2. log⁡7(343x)\log_7(343x)
  3. log⁡7(42x)\log_7(42x)
  4. log⁡7(7)+log⁡7(x)\log_7(7)+\log_7(x)
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: log_b(xy) = log_b(x) + log_b(y) because multiplying numbers means adding their exponents (b^m × b^n = b^(m+n)), (2) Quotient property: log_b(x/y) = log_b(x) - log_b(y) because dividing means subtracting exponents (b^m ÷ b^n = b^(m-n)), (3) Power property: log_b(xpx^p) = p·log_b(x) because raising to power means multiplying exponents ((b^m)^p = b^(mp)). These properties transform multiplication into addition, division into subtraction, and exponentiation into multiplication—powerful simplification tools! To simplify log₇(49x) - log₇(7) using properties: (1) Apply the quotient property backwards: log₇(49x) - log₇(7) = log₇(49x/7). (2) Simplify the fraction inside: 49x/7 = 7x because 49 = 7². (3) Therefore, log₇(49x) - log₇(7) = log₇(7x). This is the simplified form! We could also expand first: log₇(49x) = log₇(49) + log₇(x) = log₇(7²) + log₇(x) = 2 + log₇(x), so the expression becomes 2 + log₇(x) - 1 = 1 + log₇(x) = log₇(7) + log₇(x) = log₇(7x). Choice A correctly applies the quotient property to simplify the expression to log₇(7x). Choice B incorrectly calculates 49x/7 as 343x, but 49/7 = 7, not 343. The number 343 = 7³, which would come from multiplying, not dividing. Division reduces the power: 7²/7¹ = 7¹ = 7, not 7³! Simplification strategy using properties: (1) When you see subtraction of logs with same base, think quotient property: log(a) - log(b) = log(a/b). (2) Simplify the resulting fraction using regular algebra. (3) Alternatively, evaluate any logs of powers of the base: log₇(49) = log₇(7²) = 2. (4) Sometimes expanding then recombining gives insight into the answer. Both approaches—direct quotient property or expand-then-recombine—lead to the same simplified result!

Question 17

Use the change-of-base formula as needed and write in terms of natural logs: log⁡5(20).\log_5(20). (Recall: log⁡b(x)=log⁡c(x)log⁡c(b)\log_b(x)=\frac{\log_c(x)}{\log_c(b)}.)

  1. ln⁡(5)ln⁡(20)\dfrac{\ln(5)}{\ln(20)}
  2. ln⁡(20)ln⁡(5)\dfrac{\ln(20)}{\ln(5)} (correct answer)
  3. ln⁡(20)−ln⁡(5)\ln(20)-\ln(5)
  4. ln⁡(100)\ln(100)
Explanation: This question tests your understanding of the change-of-base formula, which is derived from logarithm properties, to rewrite logs in terms of natural logs. The change-of-base formula is log_b(x) = ln(x) / ln(b), coming from the power property and definition of logs, allowing conversion between bases. To express log_5(20) in natural logs, directly apply the formula: ln(20) / ln(5). Choice B correctly uses the change-of-base formula with the argument in the numerator and base in the denominator. Choice A swaps numerator and denominator—remember, it's log of x over log of b! Use change-of-base when the base doesn't match known values, always putting the argument's log on top. This tool opens up any log calculation—keep up the great work!

Question 18

Use the product, quotient, and power properties of logarithms to write the following as a single logarithm (base 10): 2log⁡(x)+12log⁡(y)−3log⁡(z).2\log(x) + \frac{1}{2}\log(y) - 3\log(z).

  1. log⁡(x2y1/2z3)\log\left(\frac{x^2 y^{1/2}}{z^3}\right) (correct answer)
  2. log⁡(x2y1/2z3)\log\left(x^2 y^{1/2} z^3\right)
  3. log⁡(x2yz3)\log\left(\frac{x^2 y}{z^3}\right)
  4. log⁡(x2+yz3)\log\left(\frac{x^2 + \sqrt{y}}{z^3}\right)
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you condense multiple logarithmic expressions into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: log⁡b(xy)=log⁡b(x)+log⁡b(y)\log_b(xy) = \log_b(x) + \log_b(y) because multiplying numbers means adding their exponents, (2) Quotient property: log⁡b(x/y)=log⁡b(x)−log⁡b(y)\log_b(x/y) = \log_b(x) - \log_b(y) because dividing means subtracting exponents, (3) Power property: log⁡b(xp)=plog⁡b(x)\log_b(x^p) = p \log_b(x) because raising to power means multiplying exponents. These properties transform addition into multiplication, subtraction into division, and coefficients into exponents—powerful simplification tools! For condensing 2log⁡(x)+12log⁡(y)−3log⁡(z)2\log(x) + \frac{1}{2}\log(y) - 3\log(z), reverse the properties: (1) Apply power property backwards: log⁡(x2)+log⁡(y1/2)−log⁡(z3)\log(x^2) + \log(y^{1/2}) - \log(z^3) by bringing coefficients up as exponents. (2) Apply product property backwards to the sum: log⁡(x2y1/2)−log⁡(z3)\log(x^2 y^{1/2}) - \log(z^3) because sum of logs becomes log of product. (3) Apply quotient property backwards: log⁡(x2y1/2z3)\log\left( \frac{x^2 y^{1/2}}{z^3} \right) because difference becomes quotient. Final condensed form: log⁡(x2yz3)\log\left( \frac{x^2 \sqrt{y}}{z^3} \right). Choice A correctly applies the logarithm properties in the right order to condense into a single logarithm. Choice B incorrectly ignores the 1/2 exponent on y, treating it as log⁡(y)\log(y) instead of log⁡(y)\log(\sqrt{y})—always remember to apply the power rule first to coefficients! Condensing strategy: (1) Use power property forward: coefficient in front becomes exponent (like 2log⁡(x)2\log(x) to log⁡(x2)\log(x^2)). (2) Identify additions: log⁡(a)+log⁡(b)\log(a) + \log(b) condenses to log⁡(ab)\log(a b). (3) Identify subtractions: log⁡(a)−log⁡(b)\log(a) - \log(b) condenses to log⁡(a/b)\log(a/b). (4) Combine all step-by-step into one log. Great job—keep building your skills with this reverse process!

Question 19

Use logarithm properties to evaluate without a calculator:

log⁡3(819)\log_3\left(\frac{81}{9}\right)

  1. 22 (correct answer)
  2. 11
  3. 44
  4. 33
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: log_b(xy) = log_b(x) + log_b(y) because multiplying numbers means adding their exponents (b^m × b^n = b^(m+n)), (2) Quotient property: log_b(x/y) = log_b(x) - log_b(y) because dividing means subtracting exponents (b^m ÷ b^n = b^(m-n)), (3) Power property: log_b(xpx^p) = p·log_b(x) because raising to power means multiplying exponents ((b^m)^p = b^(mp)). These properties transform multiplication into addition, division into subtraction, and exponentiation into multiplication—powerful simplification tools! To evaluate log₃(81/9) without a calculator: (1) First simplify the fraction: 81/9 = 9. (2) So log₃(81/9) = log₃(9). (3) Recognize that 9 = 3², so log₃(9) = log₃(3²) = 2. Alternatively, use the quotient property: log₃(81/9) = log₃(81) - log₃(9) = log₃(3⁴) - log₃(3²) = 4 - 2 = 2. Both methods give the same answer! Choice A correctly evaluates the logarithm as 2. Choice C incorrectly just uses the exponent from 81 = 3⁴ without considering the division by 9. The quotient 81/9 simplifies to 9 = 3², not 81 = 3⁴. You must perform the division inside the logarithm or use the quotient property correctly! Evaluation strategy using properties: (1) Simplify the argument first if possible: 81/9 = 9. (2) Express result as power of base: 9 = 3². (3) Use log_b(bnb^n) = n directly: log₃(3²) = 2. (4) Or use quotient property: log₃(81) - log₃(9) = 4 - 2 = 2. This flexibility in approach helps verify your answer—if two different methods give the same result, you can be confident it's correct!

Question 20

Use the logarithm properties log⁡b(xy)=log⁡b(x)+log⁡b(y)\log_b(xy)=\log_b(x)+\log_b(y), log⁡b(xy)=log⁡b(x)−log⁡b(y)\log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y), and log⁡b(xp)=plog⁡b(x)\log_b(x^p)=p\log_b(x) to expand: ln⁡(a4bc).\ln\left(\frac{a^4b}{\sqrt{c}}\right).

  1. ln⁡(a4)+ln⁡(b)−ln⁡(c)\ln(a^4)+\ln(b)-\ln(\sqrt{c})
  2. 4ln⁡(a)+ln⁡(b)−12ln⁡(c)4\ln(a)+\ln(b)-\frac{1}{2}\ln(c) (correct answer)
  3. 4ln⁡(a)+ln⁡(b)+12ln⁡(c)4\ln(a)+\ln(b)+\frac{1}{2}\ln(c)
  4. ln⁡(a+b)−ln⁡(c)\ln(a+b)-\ln(c)
Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences. The three logarithm properties come from exponent rules and work because logarithms ARE exponents: (1) Product property: ln(xy)=ln(x)+ln(y)ln(xy) = ln(x) + ln(y), (2) Quotient property: ln(x/y)=ln(x)−ln(y)ln(x/y) = ln(x) - ln(y), (3) Power property: ln(xp)=pln(x)ln(x^p) = p ln(x). These properties transform multiplication into addition, division into subtraction, and exponentiation into multiplication—powerful simplification tools! To expand ln((a4b)/sqrt(c)ln((a^4 b)/sqrt(c)): (1) Apply quotient property first: ln(a4b)−ln(sqrt(c))ln(a^4 b) - ln(sqrt(c)) for the fraction. (2) Apply product property: ln(a4)+ln(b)−12ln(c)ln(a^4) + ln(b) - \frac{1}{2}ln(c) since sqrt(c)=c1/2sqrt(c) = c^{1/2}, using power rule. (3) Apply power to a^4: 4ln(a)+ln(b)−12ln(c)4ln(a) + ln(b) - \frac{1}{2}ln(c). Final expanded form breaks it into simple terms. Choice B correctly applies the properties in order to fully expand. Choice A keeps exponents inside without using power rule—remember to bring them out as coefficients! Expanding strategy: (1) Start with outermost: quotient here. (2) Then products. (3) Finally powers. Keep going until fully simplified—you've got this!