Precalculus Examples

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

Step 1

Simplify each term.

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

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 $b = 1$ .

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

Step 1.4

Multiply $- \frac{1}{2} \ln (x - 1)$ .

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Step 1.4.1

Reorder $\ln (x - 1)$ and $\frac{1}{2}$ .

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

Step 1.4.2

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

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

Step 2

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

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

Step 3

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

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