Precalculus Examples

$4 \ln (2 x) + \ln (y) - \frac{1}{2} \cdot \ln (x + 1)$

Step 1

Simplify each term.

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Step 1.1

Simplify $4 \ln (2 x)$ by moving $4$ inside the logarithm.

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

Step 1.2

Apply the product rule to $2 x$ .

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

Step 1.3

Raise $2$ to the power of $4$ .

$\ln (16 x^{4}) + \ln (y) - \frac{1}{2} \cdot \ln (x + 1)$

Step 1.4

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

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Step 1.4.1

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

$\ln (16 x^{4}) + \ln (y) - (\frac{1}{2} \ln (x + 1))$

Step 1.4.2

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

$\ln (16 x^{4}) + \ln (y) - \ln ({(x + 1)}^{\frac{1}{2}})$

Step 2

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

$\ln (16 x^{4} y) - \ln ({(x + 1)}^{\frac{1}{2}})$

Step 3

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

$\ln (\frac{16 x^{4} y}{{(x + 1)}^{\frac{1}{2}}})$