Question

1. $log_{3} 9 = log_{2} 2$ d. $log_{3} 36$

in exercises 13–20, expand the logarithmic expression.
(see example 2.)

1. $log_{3} 4x$
2. $log_{5} 3x$
3. $log 10x^{3}$
4. $ln 3x^{4}$
5. $ln \frac{x}{3y}$
6. $ln \frac{6x^{2}}{y^{4}}$
7. $log_{2} 5sqrt3{x}$
8. $log_{5} sqrt3{x^{2}y}$

Explanation:

Step1: Apply product rule: $\log_b(xy)=\log_b x+\log_b y$

$\log_6 4x = \log_6 4 + \log_6 x$

Step2: Rewrite 4 as $2^2$

$\log_6 4 + \log_6 x = \log_6 2^2 + \log_6 x$

Step3: Apply power rule: $\log_b x^n=n\log_b x$

$\log_6 2^2 + \log_6 x = 2\log_6 2 + \log_6 x$

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Step1: Apply product rule: $\log_b(xy)=\log_b x+\log_b y$

$\log_8 3x = \log_8 3 + \log_8 x$

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Step1: Apply product rule: $\log_b(xy)=\log_b x+\log_b y$

$\log 10x^5 = \log 10 + \log x^5$

Step2: Evaluate $\log 10=1$, apply power rule

$\log 10 + \log x^5 = 1 + 5\log x$

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Step1: Apply product rule: $\log_b(xy)=\log_b x+\log_b y$

$\ln 3x^4 = \ln 3 + \ln x^4$

Step2: Apply power rule: $\log_b x^n=n\log_b x$

$\ln 3 + \ln x^4 = \ln 3 + 4\ln x$

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Step1: Apply quotient rule: $\log_b\frac{x}{y}=\log_b x-\log_b y$

$\ln \frac{x}{3y} = \ln x - \ln(3y)$

Step2: Apply product rule to $\ln(3y)$

$\ln x - \ln(3y) = \ln x - (\ln 3 + \ln y)$

Step3: Simplify the expression

$\ln x - (\ln 3 + \ln y) = \ln x - \ln 3 - \ln y$

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Step1: Apply quotient rule: $\log_b\frac{x}{y}=\log_b x-\log_b y$

$\ln \frac{6x^2}{y} = \ln 6x^2 - \ln y$

Step2: Apply product rule to $\ln 6x^2$

$\ln 6x^2 - \ln y = \ln 6 + \ln x^2 - \ln y$

Step3: Apply power rule to $\ln x^2$

$\ln 6 + \ln x^2 - \ln y = \ln 6 + 2\ln x - \ln y$

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Step1: Rewrite $\sqrt{x}$ as $x^\frac{1}{2}$

$\log_7 5\sqrt{x} = \log_7 5x^\frac{1}{2}$

Step2: Apply product rule: $\log_b(xy)=\log_b x+\log_b y$

$\log_7 5x^\frac{1}{2} = \log_7 5 + \log_7 x^\frac{1}{2}$

Step3: Apply power rule: $\log_b x^n=n\log_b x$

$\log_7 5 + \log_7 x^\frac{1}{2} = \log_7 5 + \frac{1}{2}\log_7 x$

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Step1: Rewrite $\sqrt[3]{x^2 y}$ as $(x^2 y)^\frac{1}{3}$

$\log_5 \sqrt[3]{x^2 y} = \log_5 (x^2 y)^\frac{1}{3}$

Step2: Apply power rule: $\log_b x^n=n\log_b x$

$\log_5 (x^2 y)^\frac{1}{3} = \frac{1}{3}\log_5 (x^2 y)$

Step3: Apply product rule: $\log_b(xy)=\log_b x+\log_b y$

$\frac{1}{3}\log_5 (x^2 y) = \frac{1}{3}(\log_5 x^2 + \log_5 y)$

Step4: Apply power rule to $\log_5 x^2$, distribute $\frac{1}{3}$

$\frac{1}{3}(\log_5 x^2 + \log_5 y) = \frac{2}{3}\log_5 x + \frac{1}{3}\log_5 y$

Answer:

1. $2\log_6 2 + \log_6 x$
2. $\log_8 3 + \log_8 x$
3. $1 + 5\log x$
4. $\ln 3 + 4\ln x$
5. $\ln x - \ln 3 - \ln y$
6. $\ln 6 + 2\ln x - \ln y$
7. $\log_7 5 + \frac{1}{2}\log_7 x$
8. $\frac{2}{3}\log_5 x + \frac{1}{3}\log_5 y$