Precalculus Examples

$\frac{\ln (2^{- 1} \cdot 3^{- 4})}{\ln ({((3) (5))}^{- 4})}$

Step 1

Simplify the numerator.

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Step 1.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{\ln (\frac{1}{2} \cdot 3^{- 4})}{\ln ({((3) (5))}^{- 4})}$

Step 1.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{\ln (\frac{1}{2} \cdot \frac{1}{3^{4}})}{\ln ({((3) (5))}^{- 4})}$

Step 1.3

Combine.

$\frac{\ln (\frac{1 \cdot 1}{2 \cdot 3^{4}})}{\ln ({((3) (5))}^{- 4})}$

Step 1.4

Multiply $1$ by $1$ .

$\frac{\ln (\frac{1}{2 \cdot 3^{4}})}{\ln ({((3) (5))}^{- 4})}$

Step 1.5

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

$\frac{\ln (\frac{1}{2 \cdot 81})}{\ln ({((3) (5))}^{- 4})}$

Step 1.6

Multiply $2$ by $81$ .

$\frac{\ln (\frac{1}{162})}{\ln ({((3) (5))}^{- 4})}$

Step 2

Simplify the denominator.

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Step 2.1

Multiply $3$ by $5$ .

$\frac{\ln (\frac{1}{162})}{\ln (15^{- 4})}$

Step 2.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{\ln (\frac{1}{162})}{\ln (\frac{1}{15^{4}})}$

Step 2.3

Raise $15$ to the power of $4$ .

$\frac{\ln (\frac{1}{162})}{\ln (\frac{1}{50625})}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$\frac{\ln (\frac{1}{162})}{\ln (\frac{1}{50625})}$

Decimal Form:

$0.46967337 \dots$