Precalculus Examples

$\frac{1}{4} \cdot (3 \ln (x + 2) - \ln (x) - \ln (x^{2} - 81))$

Step 1

Simplify $3 \ln (x + 2)$ by moving $3$ inside the logarithm.

$\frac{1}{4} \cdot (\ln ({(x + 2)}^{3}) - \ln (x) - \ln (x^{2} - 81))$

Step 2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{1}{4} \cdot (\ln (\frac{{(x + 2)}^{3}}{x}) - \ln (x^{2} - 81))$

Step 3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\frac{1}{4} \cdot \ln (\frac{\frac{{(x + 2)}^{3}}{x}}{x^{2} - 81})$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$\frac{1}{4} \cdot \ln (\frac{{(x + 2)}^{3}}{x} \cdot \frac{1}{x^{2} - 81})$

Step 5

Simplify the denominator.

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Step 5.1

Rewrite $81$ as $9^{2}$ .

$\frac{1}{4} \cdot \ln (\frac{{(x + 2)}^{3}}{x} \cdot \frac{1}{x^{2} - 9^{2}})$

Step 5.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 9$ .

$\frac{1}{4} \cdot \ln (\frac{{(x + 2)}^{3}}{x} \cdot \frac{1}{(x + 9) (x - 9)})$

Step 6

Combine fractions.

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Step 6.1

Combine.

$\frac{1}{4} \cdot \ln (\frac{{(x + 2)}^{3} \cdot 1}{x ((x + 9) (x - 9))})$

Step 6.2

Multiply ${(x + 2)}^{3}$ by $1$ .

$\frac{1}{4} \cdot \ln (\frac{{(x + 2)}^{3}}{x ((x + 9) (x - 9))})$

Step 7

Simplify $\frac{1}{4} \ln (\frac{{(x + 2)}^{3}}{x (x + 9) (x - 9)})$ by moving $\frac{1}{4}$ inside the logarithm.

$\ln ({(\frac{{(x + 2)}^{3}}{x (x + 9) (x - 9)})}^{\frac{1}{4}})$

Step 8

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 8.1

Apply the product rule to $\frac{{(x + 2)}^{3}}{x (x + 9) (x - 9)}$ .

$\ln (\frac{{({(x + 2)}^{3})}^{\frac{1}{4}}}{{(x (x + 9) (x - 9))}^{\frac{1}{4}}})$

Step 8.2

Apply the product rule to $x (x + 9) (x - 9)$ .

$\ln (\frac{{({(x + 2)}^{3})}^{\frac{1}{4}}}{{(x (x + 9))}^{\frac{1}{4}} {(x - 9)}^{\frac{1}{4}}})$

Step 8.3

Apply the product rule to $x (x + 9)$ .

$\ln (\frac{{({(x + 2)}^{3})}^{\frac{1}{4}}}{x^{\frac{1}{4}} {(x + 9)}^{\frac{1}{4}} {(x - 9)}^{\frac{1}{4}}})$

Step 9

Multiply the exponents in ${({(x + 2)}^{3})}^{\frac{1}{4}}$ .

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Step 9.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\ln (\frac{{(x + 2)}^{3 (\frac{1}{4})}}{x^{\frac{1}{4}} {(x + 9)}^{\frac{1}{4}} {(x - 9)}^{\frac{1}{4}}})$

Step 9.2

Combine $3$ and $\frac{1}{4}$ .

$\ln (\frac{{(x + 2)}^{\frac{3}{4}}}{x^{\frac{1}{4}} {(x + 9)}^{\frac{1}{4}} {(x - 9)}^{\frac{1}{4}}})$