Precalculus Examples

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

Step 1

Simplify each term.

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

Simplify $4 \ln (3 x)$ by moving $4$ inside the logarithm.

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

Step 1.4

Apply the product rule to $3 x$ .

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

Step 1.5

Raise $3$ to the power of $4$ .

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

Step 2

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

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

Step 3

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

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

Step 4

Rewrite using the commutative property of multiplication.

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

Step 5

Combine $81$ and $\frac{{(x - 1)}^{\frac{1}{2}}}{x^{3}}$ .

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

Step 6

Cancel the common factor of $x^{3}$ .

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Step 6.1

Factor $x^{3}$ out of $x^{4}$ .

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

Step 6.2

Cancel the common factor.

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

Step 6.3

Rewrite the expression.

$\ln (81 {(x - 1)}^{\frac{1}{2}} x)$

Step 7

Reorder factors in $\ln (81 {(x - 1)}^{\frac{1}{2}} x)$ .

$\ln (81 x {(x - 1)}^{\frac{1}{2}})$