Precalculus Examples

$(\log_{4} (12^{3})) (\log_{12} (4^{3}))$

Step 1

Raise $12$ to the power of $3$ .

$\log_{4} (1728) \log_{12} (4^{3})$

Step 2

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

$\log_{4} (1728) \log_{12} (64)$

Step 3

Rewrite $\log_{4} (1728)$ using the change of base formula.

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Step 3.1

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 $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} \log_{12} (64)$

Step 3.2

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

$\frac{\log (1728)}{\log (4)} \log_{12} (64)$

Step 4

Rewrite $\log_{12} (64)$ using the change of base formula.

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Step 4.1

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 $0$ .

$\frac{\log (1728)}{\log (4)} \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}$

Step 4.2

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

$\frac{\log (1728)}{\log (4)} \cdot \frac{\log (64)}{\log (12)}$

Step 5

Multiply $\frac{\log (1728)}{\log (4)}$ by $\frac{\log (64)}{\log (12)}$ .

$\frac{\log (1728) \log (64)}{\log (4) \log (12)}$

Step 6

Rewrite $\log (1728)$ as $\log (2^{6} \cdot 3^{3})$ .

$\frac{\log (2^{6} \cdot 3^{3}) \log (64)}{\log (4) \log (12)}$

Step 7

Rewrite $\log (2^{6} \cdot 3^{3})$ as $\log (2^{6}) + \log (3^{3})$ .

$\frac{(\log (2^{6}) + \log (3^{3})) \log (64)}{\log (4) \log (12)}$

Step 8

Rewrite $\log (64)$ as $\log (2^{6})$ .

$\frac{(\log (2^{6}) + \log (3^{3})) \log (2^{6})}{\log (4) \log (12)}$

Step 9

Expand $\log (2^{6})$ by moving $6$ outside the logarithm.

$\frac{(\log (2^{6}) + \log (3^{3})) (6 \log (2))}{\log (4) \log (12)}$

Step 10

Rewrite $\log (4)$ as $\log (2^{2})$ .

$\frac{(\log (2^{6}) + \log (3^{3})) (6 \log (2))}{\log (2^{2}) \log (12)}$

Step 11

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

$\frac{(\log (2^{6}) + \log (3^{3})) (6 \log (2))}{2 \log (2) \log (12)}$

Step 12

Rewrite $\log (12)$ as $\log (2^{2} \cdot 3)$ .

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

Step 13

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

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

Step 14

Reduce the expression by cancelling the common factors.

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Step 14.1

Cancel the common factor of $6$ and $2$ .

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Step 14.1.1

Factor $2$ out of $(\log (2^{6}) + \log (3^{3})) (6 \log (2))$ .

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

Step 14.1.2

Cancel the common factors.

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Step 14.1.2.1

Factor $2$ out of $2 \log (2) (\log (2^{2}) + \log (3))$ .

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

Step 14.1.2.2

Cancel the common factor.

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

Step 14.1.2.3

Rewrite the expression.

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

Step 14.2

Cancel the common factor of $\log (2)$ .

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Step 14.2.1

Cancel the common factor.

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

Step 14.2.2

Rewrite the expression.

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

Step 15

Simplify the numerator.

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Step 15.1

Raise $2$ to the power of $6$ .

$\frac{(\log (64) + \log (3^{3})) \cdot 3}{\log (2^{2}) + \log (3)}$

Step 15.2

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

$\frac{(\log (64) + \log (27)) \cdot 3}{\log (2^{2}) + \log (3)}$

Step 15.3

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

$\frac{\log (64 \cdot 27) \cdot 3}{\log (2^{2}) + \log (3)}$

Step 15.4

Multiply $64$ by $27$ .

$\frac{\log (1728) \cdot 3}{\log (2^{2}) + \log (3)}$

Step 16

Simplify the denominator.

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Step 16.1

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

$\frac{\log (1728) \cdot 3}{\log (4) + \log (3)}$

Step 16.2

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

$\frac{\log (1728) \cdot 3}{\log (4 \cdot 3)}$

Step 16.3

Multiply $4$ by $3$ .

$\frac{\log (1728) \cdot 3}{\log (12)}$

Step 17

Simplify the numerator.

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Step 17.1

Reorder $\log (1728)$ and $3$ .

$\frac{3 \cdot \log (1728)}{\log (12)}$

Step 17.2

Simplify $3 \cdot \log (1728)$ by moving $3$ inside the logarithm.

$\frac{\log (1728^{3})}{\log (12)}$

Step 18

Raise $1728$ to the power of $3$ .

$\frac{\log (5159780352)}{\log (12)}$

Step 19

The result can be shown in multiple forms.

Exact Form:

$\frac{\log (5159780352)}{\log (12)}$

Decimal Form:

$9$