Precalculus Examples

$\frac{\log_{9} (25)}{\log_{9} (6)}$

Step 1

Simplify the numerator.

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Step 1.1

Rewrite $\log_{9} (25)$ using the change of base formula.

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Step 1.1.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$\frac{\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}}{\log_{9} (6)}$

Step 1.1.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$\frac{\frac{\log (25)}{\log (9)}}{\log_{9} (6)}$

Step 1.2

Rewrite $\log (25)$ as $\log (5^{2})$ .

$\frac{\frac{\log (5^{2})}{\log (9)}}{\log_{9} (6)}$

Step 1.3

Expand $\log (5^{2})$ by moving $2$ outside the logarithm.

$\frac{\frac{2 \log (5)}{\log (9)}}{\log_{9} (6)}$

Step 1.4

Rewrite $\log (9)$ as $\log (3^{2})$ .

$\frac{\frac{2 \log (5)}{\log (3^{2})}}{\log_{9} (6)}$

Step 1.5

Expand $\log (3^{2})$ by moving $2$ outside the logarithm.

$\frac{\frac{2 \log (5)}{2 \log (3)}}{\log_{9} (6)}$

Step 1.6

Cancel the common factor of $2$ .

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Step 1.6.1

Cancel the common factor.

$\frac{\frac{2 \log (5)}{2 \log (3)}}{\log_{9} (6)}$

Step 1.6.2

Rewrite the expression.

$\frac{\frac{\log (5)}{\log (3)}}{\log_{9} (6)}$

Step 2

Rewrite $\log_{9} (6)$ using the change of base formula.

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Step 2.1

The change of base rule can be used if $a$ and $b$ are greater than $0$ and not equal to $1$ , and $x$ is greater than $0$ .

$\frac{\frac{\log (5)}{\log (3)}}{\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}}$

Step 2.2

Substitute in values for the variables in the change of base formula, using $b = 10$ .

$\frac{\frac{\log (5)}{\log (3)}}{\frac{\log (6)}{\log (9)}}$

Step 3

Multiply the numerator by the reciprocal of the denominator.

$\frac{\log (5)}{\log (3)} \cdot \frac{\log (9)}{\log (6)}$

Step 4

Multiply $\frac{\log (5)}{\log (3)}$ by $\frac{\log (9)}{\log (6)}$ .

$\frac{\log (5) \log (9)}{\log (3) \log (6)}$

Step 5

Rewrite $\log (9)$ as $\log (3^{2})$ .

$\frac{\log (5) \log (3^{2})}{\log (3) \log (6)}$

Step 6

Expand $\log (3^{2})$ by moving $2$ outside the logarithm.

$\frac{\log (5) (2 \log (3))}{\log (3) \log (6)}$

Step 7

Rewrite $\log (6)$ as $\log (2 (3))$ .

$\frac{\log (5) (2 \log (3))}{\log (3) \log (2 (3))}$

Step 8

Rewrite $\log (2 (3))$ as $\log (2) + \log (3)$ .

$\frac{\log (5) (2 \log (3))}{\log (3) (\log (2) + \log (3))}$

Step 9

Cancel the common factor of $\log (3)$ .

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Step 9.1

Cancel the common factor.

$\frac{\log (5) (2 \log (3))}{\log (3) (\log (2) + \log (3))}$

Step 9.2

Rewrite the expression.

$\frac{\log (5) \cdot (2)}{\log (2) + \log (3)}$

Step 10

Simplify the numerator.

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Step 10.1

Reorder $\log (5)$ and $2$ .

$\frac{2 \cdot \log (5)}{\log (2) + \log (3)}$

Step 10.2

Simplify $2 \cdot \log (5)$ by moving $2$ inside the logarithm.

$\frac{\log (5^{2})}{\log (2) + \log (3)}$

Step 11

Simplify the denominator.

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Step 11.1

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

$\frac{\log (5^{2})}{\log (2 \cdot 3)}$

Step 11.2

Multiply $2$ by $3$ .

$\frac{\log (5^{2})}{\log (6)}$

Step 12

Raise $5$ to the power of $2$ .

$\frac{\log (25)}{\log (6)}$

Step 13

The result can be shown in multiple forms.

Exact Form:

$\frac{\log (25)}{\log (6)}$

Decimal Form:

$1.79648880 \dots$