Precalculus Examples

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

Step 1

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

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

Step 2

Simplify each term.

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Combine $2$ and $\frac{1}{3}$ .

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

Step 2.4

Simplify the denominator.

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Step 2.4.1

Rewrite $1$ as $1^{2}$ .

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

Step 2.4.2

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 $b = 1$ .

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

Step 3

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 4

Combine ${(x + 3)}^{\frac{2}{3}}$ and $\frac{x}{(x + 1) (x - 1)}$ .

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

Step 5

Reorder factors in $\ln (\frac{{(x + 3)}^{\frac{2}{3}} x}{(x + 1) (x - 1)})$ .

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