College Algebra Quiz: Change Of Base
20 questions · exam conditions
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Change Of BaseQuestion 1 of 20

How can the change of base formula be used to rewrite log⁡3(5)\log_3(5) in terms of log⁡10\log_{10}?

log⁡3(5)=log⁡(3)log⁡(5)\log_3(5)=\dfrac{\log(3)}{\log(5)}
log⁡3(5)=log⁡(5)log⁡(3)\log_3(5)=\dfrac{\log(5)}{\log(3)}
log⁡3(5)=log⁡(5)−log⁡(3)\log_3(5)=\log(5)-\log(3)
log⁡3(5)=log⁡(5)log⁡(10)\log_3(5)=\dfrac{\log(5)}{\log(10)}
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College Algebra Quiz

College Algebra Quiz: Change Of Base

Practice Change Of Base 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 Change Of Base, 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

How can the change of base formula be used to rewrite log⁡3(5)\log_3(5) in terms of log⁡10\log_{10}?

  1. log⁡3(5)=log⁡(3)log⁡(5)\log_3(5)=\dfrac{\log(3)}{\log(5)}
  2. log⁡3(5)=log⁡(5)log⁡(3)\log_3(5)=\dfrac{\log(5)}{\log(3)} (correct answer)
  3. log⁡3(5)=log⁡(5)−log⁡(3)\log_3(5)=\log(5)-\log(3)
  4. log⁡3(5)=log⁡(5)log⁡(10)\log_3(5)=\dfrac{\log(5)}{\log(10)}
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_3(5) = log(5) / log(3) is used with common logs. The correct choice B is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like A fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 2

Which of the following expressions correctly uses the change of base formula to evaluate log⁡5(2)\log_5(2) with ln⁡\ln?

  1. log⁡5(2)=ln⁡(5)ln⁡(2)\log_5(2)=\dfrac{\ln(5)}{\ln(2)}
  2. log⁡5(2)=ln⁡(2)ln⁡(5)\log_5(2)=\dfrac{\ln(2)}{\ln(5)} (correct answer)
  3. log⁡5(2)=ln⁡(2)−ln⁡(5)\log_5(2)=\ln(2)-\ln(5)
  4. log⁡5(2)=ln⁡(2) ln⁡(5)\log_5(2)=\ln(2)\,\ln(5)
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_5(2) = ln(2) / ln(5) is used with natural logs. The correct choice B is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like A fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 3

Which of the following expressions correctly uses the change of base formula to evaluate log⁡4(9)\log_4(9) with ln⁡\ln?

  1. log⁡4(9)=ln⁡(4)ln⁡(9)\log_4(9)=\dfrac{\ln(4)}{\ln(9)}
  2. log⁡4(9)=ln⁡(9)ln⁡(4)\log_4(9)=\dfrac{\ln(9)}{\ln(4)} (correct answer)
  3. log⁡4(9)=ln⁡(9)−ln⁡(4)\log_4(9)=\ln(9)-\ln(4)
  4. log⁡4(9)=ln⁡(9) ln⁡(4)\log_4(9)=\ln(9)\,\ln(4)
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_4(9) = ln(9) / ln(4) is used with natural logs. The correct choice B is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like A fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 4

If log⁡2(5)=a\log_2(5) = a and log⁡3(5)=b\log_3(5) = b, then log⁡6(5)\log_6(5) can be expressed in terms of aa and bb. Using the change of base formula and properties of logarithms, what is this expression?

  1. 1a+1b\frac{1}{a} + \frac{1}{b}
  2. a+bab\frac{a+b}{ab}
  3. aba+b\frac{ab}{a+b} (correct answer)
  4. 2aba+b\frac{2ab}{a+b}
Explanation: When you encounter logarithms with different bases that need to be expressed in terms of given logarithmic values, the change of base formula is your primary tool. This formula states that log⁡c(x)=log⁡a(x)log⁡a(c)\log_c(x) = \frac{\log_a(x)}{\log_a(c)} for any valid base aa. To find log⁡6(5)\log_6(5), you need to express it using bases 2 and 3 since you're given log⁡2(5)=a\log_2(5) = a and log⁡3(5)=b\log_3(5) = b. Using the change of base formula with base 2: log⁡6(5)=log⁡2(5)log⁡2(6)\log_6(5) = \frac{\log_2(5)}{\log_2(6)}. Since 6=2×36 = 2 \times 3, you can use the logarithm property log⁡(xy)=log⁡(x)+log⁡(y)\log(xy) = \log(x) + \log(y) to get: log⁡2(6)=log⁡2(2)+log⁡2(3)=1+log⁡2(3)\log_2(6) = \log_2(2) + \log_2(3) = 1 + \log_2(3). To find log⁡2(3)\log_2(3), use the change of base formula again: log⁡2(3)=1log⁡3(2)\log_2(3) = \frac{1}{\log_3(2)}. Since log⁡3(5)=b\log_3(5) = b and using properties of logarithms, you can show that log⁡2(3)=1log⁡3(2)=ab\log_2(3) = \frac{1}{\log_3(2)} = \frac{a}{b} through the relationship log⁡2(3)=log⁡3(5)log⁡3(2)⋅1log⁡2(5)=b1/a=ab\log_2(3) = \frac{\log_3(5)}{\log_3(2)} \cdot \frac{1}{\log_2(5)} = \frac{b}{1/a} = \frac{a}{b}. Wait, let me recalculate more directly: log⁡6(5)=log⁡2(5)log⁡2(6)=a1+1log⁡3(2)=a1+ab=aba+b\log_6(5) = \frac{\log_2(5)}{\log_2(6)} = \frac{a}{1 + \frac{1}{\log_3(2)}} = \frac{a}{1 + \frac{a}{b}} = \frac{ab}{a+b} Answer choice A gives 1a+1b\frac{1}{a} + \frac{1}{b}, which incorrectly adds reciprocals. Choice B gives a+bab\frac{a+b}{ab}, which is the reciprocal of our answer. Choice D includes an extra factor of 2 that doesn't belong. Choice C, aba+b\frac{ab}{a+b}, is correct. Strategy tip: When working with change of base problems, systematically convert everything to the given bases and remember that log⁡a(bc)=log⁡a(b)+log⁡a(c)\log_a(bc) = \log_a(b) + \log_a(c).

Question 5

Given log⁡4(x)=12\log_4(x)=\dfrac{1}{2}, which change-of-base equation in ln⁡\ln is equivalent?

  1. ln⁡(4)ln⁡(x)=12\dfrac{\ln(4)}{\ln(x)}=\dfrac{1}{2}
  2. ln⁡(x)ln⁡(4)=12\dfrac{\ln(x)}{\ln(4)}=\dfrac{1}{2} (correct answer)
  3. ln⁡(x)−ln⁡(4)=12\ln(x)-\ln(4)=\dfrac{1}{2}
  4. ln⁡(x) ln⁡(4)=12\ln(x)\,\ln(4)=\dfrac{1}{2}
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_4(x) = ln(x) / ln(4) = 1/2 is used with natural logs. The correct choice B is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like A fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 6

Given log⁡8(4)=x\log_8(4)=x, use the change of base formula to express xx in terms of ln⁡\ln.

  1. x=ln⁡(8)ln⁡(4)x=\dfrac{\ln(8)}{\ln(4)}
  2. x=ln⁡(4)ln⁡(8)x=\dfrac{\ln(4)}{\ln(8)} (correct answer)
  3. x=ln⁡(4)ln⁡(10)x=\dfrac{\ln(4)}{\ln(10)}
  4. x=ln⁡(4)−ln⁡(8)x=\ln(4)-\ln(8)
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_8(4) = ln(4) / ln(8) is used with natural logs. The correct choice B is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like A fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 7

To solve the equation 2x=3x−12^x = 3^{x-1}, a student applies natural logarithm to both sides and uses the change of base concept. Which of the following correctly represents the next step in the solution process?

  1. xln⁡(2)=(x−1)ln⁡(3)x \ln(2) = (x-1) \ln(3) (correct answer)
  2. ln⁡(2)x=ln⁡(3)x−1\frac{\ln(2)}{x} = \frac{\ln(3)}{x-1}
  3. x=ln⁡(3)ln⁡(2)⋅(x−1)x = \frac{\ln(3)}{\ln(2)} \cdot (x-1)
  4. ln⁡(x)⋅2=ln⁡(x−1)⋅3\ln(x) \cdot 2 = \ln(x-1) \cdot 3
Explanation: Taking the natural logarithm of both sides: ln⁡(2x)=ln⁡(3x−1)\ln(2^x) = \ln(3^{x-1}). Using the power rule for logarithms: xln⁡(2)=(x−1)ln⁡(3)x\ln(2) = (x-1)\ln(3). This can then be solved for x by expanding and collecting terms. Choice B incorrectly applies the change of base formula where it doesn't belong. Choice C prematurely tries to isolate x without properly applying the logarithm to the exponential equation. Choice D incorrectly applies the logarithm to the exponents rather than the entire exponential expressions.

Question 8

The equation log⁡4(x−1)=log⁡7(x+3)\log_4(x-1) = \log_7(x+3) can be solved by converting both sides to the same base. If you convert both logarithms to natural logarithms using the change of base formula, which equation results?

  1. ln⁡(x−1)ln⁡(4)=ln⁡(x+3)ln⁡(7)\frac{\ln(x-1)}{\ln(4)} = \frac{\ln(x+3)}{\ln(7)} (correct answer)
  2. ln⁡(4)ln⁡(x−1)=ln⁡(7)ln⁡(x+3)\frac{\ln(4)}{\ln(x-1)} = \frac{\ln(7)}{\ln(x+3)}
  3. ln⁡(4)⋅ln⁡(x−1)=ln⁡(7)⋅ln⁡(x+3)\ln(4) \cdot \ln(x-1) = \ln(7) \cdot \ln(x+3)
  4. ln⁡(x−1)ln⁡(7)=ln⁡(x+3)ln⁡(4)\frac{\ln(x-1)}{\ln(7)} = \frac{\ln(x+3)}{\ln(4)}
Explanation: The change of base formula states that log⁡ba=ln⁡aln⁡b\log_b a = \frac{\ln a}{\ln b}. Applying this to both sides: log⁡4(x−1)=ln⁡(x−1)ln⁡(4)\log_4(x-1) = \frac{\ln(x-1)}{\ln(4)} and log⁡7(x+3)=ln⁡(x+3)ln⁡(7)\log_7(x+3) = \frac{\ln(x+3)}{\ln(7)}. Therefore, the equation becomes ln⁡(x−1)ln⁡(4)=ln⁡(x+3)ln⁡(7)\frac{\ln(x-1)}{\ln(4)} = \frac{\ln(x+3)}{\ln(7)}. Choice B flips the numerator and denominator. Choice C incorrectly uses multiplication instead of division in the change of base formula. Choice D switches the bases in the denominators.

Question 9

A scientific calculator displays log⁡2(50)≈5.644\log_2(50) \approx 5.644. Using this information and the change of base formula, what is the approximate value of log⁡50(2)\log_{50}(2)?

  1. 0.1770.177 (correct answer)
  2. −5.644-5.644
  3. 5.6445.644
  4. 0.6440.644
Explanation: Using the change of base formula, log⁡50(2)=log⁡2(2)log⁡2(50)=1log⁡2(50)=15.644≈0.177\log_{50}(2) = \frac{\log_2(2)}{\log_2(50)} = \frac{1}{\log_2(50)} = \frac{1}{5.644} \approx 0.177. This demonstrates that log⁡a(b)\log_a(b) and log⁡b(a)\log_b(a) are reciprocals when both are positive. Choice B would be incorrect as it suggests a negative logarithm when both base and argument are greater than 1. Choice C incorrectly assumes the logarithms are equal. Choice D appears to subtract 5 from 5.644, which has no mathematical basis.

Question 10

Given log⁡5(25)=x\log_5(25)=x, use the change of base formula to express xx in terms of log⁡10\log_{10}.

  1. x=log⁡(25)log⁡(5)x=\dfrac{\log(25)}{\log(5)} (correct answer)
  2. x=log⁡(5)log⁡(25)x=\dfrac{\log(5)}{\log(25)}
  3. x=log⁡(25)log⁡(10)x=\dfrac{\log(25)}{\log(10)}
  4. x=log⁡(25)−log⁡(5)x=\log(25)-\log(5)
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_5(25) = log(25) / log(5) is used with common logs. The correct choice A is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like B fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 11

Which of the following expressions correctly uses the change of base formula to evaluate log⁡12(18)\log_{12}(18) with ln⁡\ln?

  1. log⁡12(18)=ln⁡(12)ln⁡(18)\log_{12}(18)=\dfrac{\ln(12)}{\ln(18)}
  2. log⁡12(18)=ln⁡(18)ln⁡(12)\log_{12}(18)=\dfrac{\ln(18)}{\ln(12)} (correct answer)
  3. log⁡12(18)=ln⁡(18)ln⁡(10)\log_{12}(18)=\dfrac{\ln(18)}{\ln(10)}
  4. log⁡12(18)=ln⁡(18)−ln⁡(12)\log_{12}(18)=\ln(18)-\ln(12)
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_12(18) = ln(18) / ln(12) is used with natural logs. The correct choice B is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like A fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 12

How can the change of base formula be used to rewrite log⁡7(49)\log_7(49) in terms of common logarithms?

  1. log⁡7(49)=log⁡(7)log⁡(49)\log_7(49)=\dfrac{\log(7)}{\log(49)}
  2. log⁡7(49)=log⁡(49)log⁡(7)\log_7(49)=\dfrac{\log(49)}{\log(7)} (correct answer)
  3. log⁡7(49)=log⁡(49)log⁡(10)\log_7(49)=\dfrac{\log(49)}{\log(10)}
  4. log⁡7(49)=log⁡(49)+log⁡(7)\log_7(49)=\log(49)+\log(7)
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_7(49) = log(49) / log(7) is used with common logs. The correct choice B is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like A fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 13

Which of the following expressions correctly uses the change of base formula to evaluate log⁡7(2)\log_7(2) with log⁡\log?

  1. log⁡7(2)=log⁡(2)log⁡(7)\log_7(2)=\dfrac{\log(2)}{\log(7)} (correct answer)
  2. log⁡7(2)=log⁡(7)log⁡(2)\log_7(2)=\dfrac{\log(7)}{\log(2)}
  3. log⁡7(2)=log⁡(2)−log⁡(7)\log_7(2)=\log(2)-\log(7)
  4. log⁡7(2)=log⁡(2)log⁡(10)\log_7(2)=\dfrac{\log(2)}{\log(10)}
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_7(2) = log(2) / log(7) is used with common logs. The correct choice A is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like B fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 14

How can the change of base formula be used to rewrite log⁡2(32)\log_2(32) in terms of natural logarithms?

  1. log⁡2(32)=ln⁡(2)ln⁡(32)\log_2(32)=\dfrac{\ln(2)}{\ln(32)}
  2. log⁡2(32)=ln⁡(32)ln⁡(2)\log_2(32)=\dfrac{\ln(32)}{\ln(2)} (correct answer)
  3. log⁡2(32)=ln⁡(32)ln⁡(10)\log_2(32)=\dfrac{\ln(32)}{\ln(10)}
  4. log⁡2(32)=ln⁡(32)−ln⁡(2)\log_2(32)=\ln(32)-\ln(2)
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_2(32) = ln(32) / ln(2) is used with natural logs. The correct choice B is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like A fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 15

How can the change of base formula be used to rewrite log⁡16(2)\log_{16}(2) in terms of common logarithms?

  1. log⁡16(2)=log⁡(16)log⁡(2)\log_{16}(2)=\dfrac{\log(16)}{\log(2)}
  2. log⁡16(2)=log⁡(2)log⁡(16)\log_{16}(2)=\dfrac{\log(2)}{\log(16)} (correct answer)
  3. log⁡16(2)=log⁡(2)−log⁡(16)\log_{16}(2)=\log(2)-\log(16)
  4. log⁡16(2)=log⁡(2)log⁡(10)\log_{16}(2)=\dfrac{\log(2)}{\log(10)}
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_{16}(2) = log(2) / log(16) is used with common logs. The correct choice B is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like A fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 16

Given log⁡3(x)=4\log_3(x)=4, which expression uses change of base to rewrite it with ln⁡\ln?

  1. ln⁡(3)ln⁡(x)=4\dfrac{\ln(3)}{\ln(x)}=4
  2. ln⁡(x)ln⁡(3)=4\dfrac{\ln(x)}{\ln(3)}=4 (correct answer)
  3. ln⁡(x)−ln⁡(3)=4\ln(x)-\ln(3)=4
  4. ln⁡(x) ln⁡(3)=4\ln(x)\,\ln(3)=4
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_3(x) = ln(x) / ln(3) = 4 is used with natural logs. The correct choice B is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like A fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 17

A student wants to evaluate log⁡0.5(8)\log_{0.5}(8) using a calculator that only has base-10 logarithms. After applying the change of base formula correctly, which numerical calculation should be performed?

  1. log⁡(8)−log⁡(0.5)≈0.903−(−0.301)≈1.204\log(8) - \log(0.5) \approx 0.903 - (-0.301) \approx 1.204
  2. log⁡(0.5)log⁡(8)≈−0.3010.903≈−0.333\frac{\log(0.5)}{\log(8)} \approx \frac{-0.301}{0.903} \approx -0.333
  3. log⁡(8)log⁡(0.5)≈0.9030.301≈3\frac{\log(8)}{\log(0.5)} \approx \frac{0.903}{0.301} \approx 3
  4. log⁡(8)log⁡(0.5)≈0.903−0.301≈−3\frac{\log(8)}{\log(0.5)} \approx \frac{0.903}{-0.301} \approx -3 (correct answer)
Explanation: When you encounter a logarithm with an unfamiliar base on a calculator that only has common logarithms (base 10), you need the change of base formula: log⁡b(a)=log⁡(a)log⁡(b)\log_b(a) = \frac{\log(a)}{\log(b)}. This formula converts any logarithm to a ratio of common logarithms. For log⁡0.5(8)\log_{0.5}(8), you're looking for the power to which 0.5 must be raised to get 8. Applying the change of base formula: log⁡0.5(8)=log⁡(8)log⁡(0.5)\log_{0.5}(8) = \frac{\log(8)}{\log(0.5)}. Now you need the actual values: log⁡(8)≈0.903\log(8) \approx 0.903 and log⁡(0.5)≈−0.301\log(0.5) \approx -0.301 (negative because 0.5 < 1). This gives you 0.903−0.301≈−3\frac{0.903}{-0.301} \approx -3. Choice A incorrectly uses subtraction instead of division—this isn't the change of base formula. Choice B has the fraction upside down, calculating log⁡8(0.5)\log_8(0.5) instead of log⁡0.5(8)\log_{0.5}(8). Choice C makes the critical error of treating log⁡(0.5)\log(0.5) as positive 0.301, when it's actually negative since 0.5 < 1. Only choice D correctly applies the formula with the proper signs. You can verify this makes sense: since the base 0.5 is between 0 and 1, and we want a result greater than 1, the logarithm should be negative. Indeed, 0.5−3=10.53=10.125=80.5^{-3} = \frac{1}{0.5^3} = \frac{1}{0.125} = 8. Study tip: Always remember that log⁡(x)\log(x) is negative when 0<x<10 < x < 1, and double-check that your change of base setup has the argument in the numerator and the new base in the denominator.

Question 18

How can the change of base formula be used to rewrite log⁡3(81)\log_3(81) in terms of natural logarithms?

  1. log⁡3(81)=ln⁡(3)ln⁡(81)\log_3(81)=\dfrac{\ln(3)}{\ln(81)}
  2. log⁡3(81)=ln⁡(81)ln⁡(3)\log_3(81)=\dfrac{\ln(81)}{\ln(3)} (correct answer)
  3. log⁡3(81)=ln⁡(81)ln⁡(10)\log_3(81)=\dfrac{\ln(81)}{\ln(10)}
  4. log⁡3(81)=ln⁡(81)−ln⁡(3)\log_3(81)=\ln(81)-\ln(3)
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_b(x) = ln(x) / ln(b) is used to express log_3(81) with natural logs. The correct choice B is valid because it applies the formula correctly by placing the logarithm of the argument in the numerator and the base in the denominator, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like A fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.

Question 19

If log⁡a(x)=3.2\log_a(x) = 3.2 and log⁡b(x)=4.8\log_b(x) = 4.8, what is the value of log⁡a(b)\log_a(b) expressed in terms of these given values?

  1. 4.83.2=1.5\frac{4.8}{3.2} = 1.5
  2. 3.24.8=23\frac{3.2}{4.8} = \frac{2}{3} (correct answer)
  3. 4.8−3.2=1.64.8 - 3.2 = 1.6
  4. 3.2+4.8=8.03.2 + 4.8 = 8.0
Explanation: Using the change of base formula: log⁡a(x)=log⁡b(x)log⁡b(a)\log_a(x) = \frac{\log_b(x)}{\log_b(a)}, so 3.2=4.8log⁡b(a)3.2 = \frac{4.8}{\log_b(a)}. Solving for log⁡b(a)\log_b(a): log⁡b(a)=4.83.2=1.5\log_b(a) = \frac{4.8}{3.2} = 1.5. Since log⁡a(b)=1log⁡b(a)\log_a(b) = \frac{1}{\log_b(a)}, we have log⁡a(b)=11.5=23\log_a(b) = \frac{1}{1.5} = \frac{2}{3}. Choice A gives log⁡b(a)\log_b(a) instead of log⁡a(b)\log_a(b). Choices C and D use addition/subtraction, which don't apply to this relationship between logarithms with different bases.

Question 20

How can the change of base formula be used to rewrite log⁡1/2(8)\log_{1/2}(8) in terms of common logarithms?

  1. log⁡1/2(8)=log⁡(1/2)log⁡(8)\log_{1/2}(8)=\dfrac{\log(1/2)}{\log(8)}
  2. log⁡1/2(8)=log⁡(8)log⁡(1/2)\log_{1/2}(8)=\dfrac{\log(8)}{\log(1/2)} (correct answer)
  3. log⁡1/2(8)=log⁡(8)log⁡(10)\log_{1/2}(8)=\dfrac{\log(8)}{\log(10)}
  4. log⁡1/2(8)=log⁡(8)−log⁡(1/2)\log_{1/2}(8)=\log(8)-\log(1/2)
Explanation: This question tests understanding of the change of base formula in logarithms and its application in simplifying complex logarithmic expressions. The change of base formula allows us to rewrite logarithms of any base in terms of common logarithms (base 10) or natural logarithms (base e), which are easier to evaluate using standard calculators. In the given question, the formula log_{1/2}(8) = log(8) / log(1/2) is used with common logs. The correct choice B is valid because it applies the formula correctly, showing an understanding of how to manipulate logarithmic expressions to a simpler form. A common distractor like A fails by swapping numerator and denominator, a typical error when students confuse the roles of the argument and base in logarithmic expressions. Teaching strategies include practicing the formula with different bases, reinforcing understanding through real-world applications like pH or decibels, and highlighting the importance of base selection in simplifying calculations.