Logarithmic Function Manipulation

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AP Precalculus › Logarithmic Function Manipulation

Questions 1 - 10
1

Based on the logarithmic expression provided, which property is used to rewrite $\log_b(M^3)$ as $3\log_b(M)$?

Quotient rule

Product rule

Change-of-base formula

Power rule

Explanation

This question tests AP Precalculus skills in logarithmic function manipulation, specifically identifying which logarithmic property allows us to move an exponent outside the logarithm. The power rule for logarithms states that log_$b(M^n$) = n×log_b(M), which allows us to bring exponents out as coefficients. In this problem, we're asked to identify the property that rewrites log_b(M³) as 3log_b(M), which is a direct application of the power rule. Choice B is correct because the power rule specifically deals with exponents inside logarithms. Choice A (quotient rule) is incorrect because it deals with log(M/N), not exponents, while choice C (product rule) handles log(M×N), and choice D (change of base) converts between different logarithmic bases. To help students, create a reference chart of all logarithmic properties with clear examples. Emphasize pattern recognition: exponents inside become coefficients outside (power rule), multiplication inside becomes addition outside (product rule), and division inside becomes subtraction outside (quotient rule).

2

Based on the logarithmic expression provided, simplify $\log_3(27x^2)-\log_3(3x)$ for $x>0$.

$\log_3!\left(\dfrac{x}{9}\right)$

$\log_3(81x^3)$

$\log_3(9x^2)$

$\log_3(9x)$

Explanation

This question tests AP Precalculus skills in logarithmic function manipulation, specifically applying the quotient rule for logarithms to simplify expressions. The quotient rule states that log_b(M) - log_b(N) = log_b(M/N), which allows us to combine logarithms with the same base through division. In this problem, we have log_3(27x²) - log_3(3x), which becomes log_3(27x²/3x) = log_3(9x) after simplifying the fraction. Choice A is correct because 27x²/3x = 9x when we divide the coefficients (27/3 = 9) and subtract the exponents of x (x²/x = x¹). Choice C is incorrect because it represents log_3(27x²) - log_3(3), missing the x in the second term. To help students, emphasize careful algebraic simplification within logarithmic expressions and practice identifying when to apply the quotient rule. Watch for errors in simplifying fractions or forgetting to include all variables.

3

Based on the logarithmic expression provided, simplify $\log_5(125a^3)+\log_5!\left(\dfrac{1}{a}\right)$ for $a>0$.

$\log_5!\left(\dfrac{125}{a^2}\right)$

$\log_5(125a^4)$

$\log_5!\left(\dfrac{a^2}{125}\right)$

$\log_5(125a^2)$

Explanation

This question tests AP Precalculus skills in logarithmic function manipulation, specifically applying the product rule for logarithms to combine expressions. The product rule states that log_b(M) + log_b(N) = log_b(M×N), allowing us to combine logarithms with the same base through multiplication. In this problem, we have log_5(125a³) + log_5(1/a), which becomes log_5(125a³ × 1/a) = log_5(125a²) after simplification. Choice A is correct because 125a³ × 1/a = 125a³⁻¹ = 125a², where we multiply the coefficients and add the exponents of a (3 + (-1) = 2). Choice B is incorrect because it would result from log_5(125a³) + log_5(a), not log_5(1/a). To help students, emphasize recognizing that 1/a = a⁻¹ and practice combining positive and negative exponents. Watch for errors in handling reciprocals and applying exponent rules within logarithmic expressions.

4

Based on the logarithmic expression provided, simplify $\log_4(16x^2)-\tfrac12\log_4(x)$ for $x>0$.

$\log_4!\left(\dfrac{16x^2}{x}\right)$

$\log_4(16x^{5/2})$

$\log_4(16x^{3/2})$

$\log_4(8x)$

Explanation

This question tests AP Precalculus skills in logarithmic function manipulation, specifically combining the quotient rule with the power rule for logarithms. The power rule states that n×log_b(M) = log_$b(M^n$), and we need to apply this before using the quotient rule. In this problem, we have log_4(16x²) - ½log_4(x), where ½log_4(x) = log_4(x^(1/2)) = log_4(√x). Choice A is correct because log_4(16x²) - log_4(√x) = log_4(16x²/√x) = log_4(16x²/x^(1/2)) = log_4(16x^(2-1/2)) = log_4(16x^(3/2)). Choice B is incorrect because it would result from adding the exponents instead of subtracting them. To help students master this concept, practice converting coefficients in front of logarithms to exponents inside, then applying quotient or product rules. Watch for errors in fractional exponent arithmetic and forgetting to apply the power rule before combining terms.

5

Based on the logarithmic expression provided, convert $\log_2!\left(\frac{I}{I_0}\right)$ to base 10 for a sound-intensity ratio.

$\dfrac{\log_{10}(2)}{\log_{10}(I/I_0)}$

$\log_{10}(I/I_0),\log_{10}(2)$

$\log_{10}(I/I_0)+\log_{10}(2)$

$\dfrac{\log_{10}(I/I_0)}{\log_{10}(2)}$

Explanation

This question tests AP Precalculus skills in logarithmic function manipulation, specifically applying the change of base formula to convert between different logarithmic bases. The change of base formula states that log_a(x) = log_b(x)/log_b(a) for any valid bases a and b. In this problem, we need to convert log_2(I/I_0) to base 10, which means applying the formula with a=2, b=10, and x=I/I_0. Choice A is correct because it properly applies the change of base formula: log_2(I/I_0) = log_10(I/I_0)/log_10(2). Choice B is incorrect because it inverts the fraction, placing log_10(2) in the numerator instead of the denominator. To help students master this concept, emphasize memorizing the change of base formula and practicing with various base conversions. Common errors include inverting the fraction or confusing which logarithm goes in the numerator versus denominator.

6

Based on the logarithmic expression provided, simplify $\log_2(8x)-\log_2(x)+\log_2!\left(\tfrac14\right)$ for $x>0$.

$\log_2(2x)$

$\log_2(2)$

$\log_2!\left(\tfrac{x}{2}\right)$

$\log_2(32)$

Explanation

This question tests AP Precalculus skills in logarithmic function manipulation, specifically applying the quotient and product rules of logarithms. Logarithmic manipulation involves using properties such as the difference of logs becoming the log of a quotient and the sum becoming the log of a product to simplify expressions. In this problem, the expression (\log_2(8x) - \log_2(x) + \log_2($\frac{1}{4}$)) requires combining terms using these rules. Choice C is correct because simplifying yields (\log_2(8) + \log_2($\frac{1}{4}$) = \log_2(2)), as the x terms cancel out. Choice A is incorrect because it fails to account for the cancellation and the addition of (\log_2($\frac{1}{4}$)), resulting in an extra factor of x. To help students, emphasize practicing the order of operations in log simplifications and verifying by exponentiating back to the original base. Watch for common errors like forgetting to apply the quotient rule properly or mishandling constants.

7

Using the provided logarithmic data, convert the sound level formula $L=10\log_{10}(I/I_0)$ into base 2 for $\log_{10}(I/I_0)$.

$\dfrac{\log_2(I/I_0)}{\log_2(10)}$

$\dfrac{\log_2(10)}{\log_2(I/I_0)}$

$\dfrac{\log_{10}(I/I_0)}{\log_{10}(2)}$

$\log_2(10),\log_2(I/I_0)$

Explanation

This question tests AP Precalculus skills in logarithmic function manipulation, specifically using the change of base formula to convert between logarithmic bases. Logarithmic manipulation involves applying the change of base formula, (\log_b(a) = $\frac{\log_k(a)}{\log_k(b)}$), to rewrite expressions in different bases. In this problem, the task is to convert $(\log_{10}$(I/I_0)) from base 10 to base 2 using the change of base formula. Choice A is correct because it accurately applies the formula, yielding ($\frac{\log_2(I/I_0)}{\log_2(10)}$). Choice B is incorrect because it swaps the bases incorrectly, using base 10 in the denominator with log base 10 in the numerator. To help students, encourage deriving the change of base formula from basic log properties and practicing conversions between common bases like 2, 10, and e. Watch for common errors like inverting the fraction or confusing the base and argument.

8

Based on the logarithmic expression provided, simplify $\log_2(8m)-\log_2!\left(\dfrac{m}{4}\right)$ for $m>0$.

$\log_2(32)$

$\log_2(32m^2)$

$\log_2!\left(\dfrac{m^2}{32}\right)$

$\log_2(2m)$

Explanation

This question tests AP Precalculus skills in logarithmic function manipulation, specifically applying the quotient rule to simplify logarithmic expressions. The quotient rule states that log_b(M) - log_b(N) = log_b(M/N), allowing us to combine logarithms through division. In this problem, we have log_2(8m) - log_2(m/4), which becomes log_2(8m ÷ m/4) = log_2(8m × 4/m) = log_2(32) after simplification. Choice A is correct because 8m ÷ (m/4) = 8m × 4/m = 32m/m = 32, where the m terms cancel out. Choice C is incorrect because it might result from misapplying the rules or making an algebraic error. To help students, emphasize that dividing by a fraction means multiplying by its reciprocal, and practice simplifying complex fractions within logarithmic expressions. Watch for errors in handling division by fractions and canceling variables correctly.

9

Based on the logarithmic expression provided, convert $\log_7(49k)$ to base 10 using change of base.

$\log_{10}(49k),\log_{10}(7)$

$\dfrac{\log_{10}(49k)}{\log_{10}(7)}$

$\log_{10}(49k)-\log_{10}(7)$

$\dfrac{\log_{10}(7)}{\log_{10}(49k)}$

Explanation

This question tests AP Precalculus skills in logarithmic function manipulation, specifically applying the change of base formula to convert from base 7 to base 10. The change of base formula states that log_a(x) = log_b(x)/log_b(a), which allows conversion between any two valid bases. In this problem, we need to convert log_7(49k) to base 10, applying the formula with a=7, b=10, and x=49k. Choice A is correct because it properly applies the change of base formula: log_7(49k) = log_10(49k)/log_10(7). Choice B is incorrect because it inverts the fraction, incorrectly placing log_10(7) in the numerator. To help students master this concept, provide a clear mnemonic for the change of base formula and practice with various conversions. Common errors include inverting the fraction or attempting to simplify 49 as 7² before applying the formula, which is unnecessary for the conversion.

10

Using the provided logarithmic data, simplify $\ln(5t)-\ln(t^2)$ for $t>0$.

$\ln!\left(\dfrac{5}{t}\right)$

$\ln!\left(\dfrac{t}{5}\right)$

$\ln!\left(\dfrac{5}{t^2}\right)$

$\ln(5t^3)$

Explanation

This question tests AP Precalculus skills in logarithmic function manipulation, specifically applying the quotient rule for natural logarithms. The quotient rule states that ln(M) - ln(N) = ln(M/N), allowing us to combine logarithms through division. In this problem, we need to simplify ln(5t) - ln(t²), which becomes ln(5t/t²) = ln(5/t) after algebraic simplification. Choice A is correct because when we divide 5t by t², we get 5t/t² = 5/t (subtracting exponents: t¹⁻² = t⁻¹ = 1/t). Choice B is incorrect because it would result from adding the logarithms instead of subtracting them, giving ln(5t × t²) = ln(5t³). To help students master this skill, emphasize the importance of recognizing subtraction as an indicator to use the quotient rule, not the product rule. Practice simplifying algebraic fractions within logarithmic expressions, watching for sign errors and incorrect exponent rules.

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