Precalculus Examples

$\frac{1}{8} \cdot (3 \ln (x + 6)) - \ln (x) - \ln (x^{2} - 49)$

Step 1

Simplify each term.

Tap for more steps...

Step 1.1

Simplify $3 \ln (x + 6)$ by moving $3$ inside the logarithm.

$\frac{1}{8} \cdot \ln ({(x + 6)}^{3}) - \ln (x) - \ln (x^{2} - 49)$

Step 1.2

Simplify $\frac{1}{8} \ln ({(x + 6)}^{3})$ by moving $\frac{1}{8}$ inside the logarithm.

$\ln ({({(x + 6)}^{3})}^{\frac{1}{8}}) - \ln (x) - \ln (x^{2} - 49)$

Step 1.3

Multiply the exponents in ${({(x + 6)}^{3})}^{\frac{1}{8}}$ .

Tap for more steps...

Step 1.3.1

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

$\ln ({(x + 6)}^{3 (\frac{1}{8})}) - \ln (x) - \ln (x^{2} - 49)$

Step 1.3.2

Combine $3$ and $\frac{1}{8}$ .

$\ln ({(x + 6)}^{\frac{3}{8}}) - \ln (x) - \ln (x^{2} - 49)$

Step 2

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

$\ln (\frac{{(x + 6)}^{\frac{3}{8}}}{x}) - \ln (x^{2} - 49)$

Step 3

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

$\ln (\frac{\frac{{(x + 6)}^{\frac{3}{8}}}{x}}{x^{2} - 49})$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$\ln (\frac{{(x + 6)}^{\frac{3}{8}}}{x} \cdot \frac{1}{x^{2} - 49})$

Step 5

Simplify the denominator.

Tap for more steps...

Step 5.1

Rewrite $49$ as $7^{2}$ .

$\ln (\frac{{(x + 6)}^{\frac{3}{8}}}{x} \cdot \frac{1}{x^{2} - 7^{2}})$

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 $b = 7$ .

$\ln (\frac{{(x + 6)}^{\frac{3}{8}}}{x} \cdot \frac{1}{(x + 7) (x - 7)})$

Step 6

Combine fractions.

Tap for more steps...

Step 6.1

Combine.

$\ln (\frac{{(x + 6)}^{\frac{3}{8}} \cdot 1}{x ((x + 7) (x - 7))})$

Step 6.2

Multiply ${(x + 6)}^{\frac{3}{8}}$ by $1$ .

$\ln (\frac{{(x + 6)}^{\frac{3}{8}}}{x (x + 7) (x - 7)})$