Precalculus Examples

$\frac{\log_{4} (81) - \log_{p} (1)}{\log_{4.24} (18) - \log (0.01)}$

Step 1

Simplify the numerator.

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Step 1.1

Logarithm base $p$ of $1$ is $0$ .

$\frac{\log_{4} (81) - 0}{\log_{4.24} (18) - \log (0.01)}$

Step 1.2

Multiply $- 1$ by $0$ .

$\frac{\log_{4} (81) + 0}{\log_{4.24} (18) - \log (0.01)}$

Step 1.3

Add $\log_{4} (81)$ and $0$ .

$\frac{\log_{4} (81)}{\log_{4.24} (18) - \log (0.01)}$

Step 2

Simplify the denominator.

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Step 2.1

Log base $10$ of $0.01$ is approximately $- 2$ .

$\frac{\log_{4} (81)}{\log_{4.24} (18) - - 2}$

Step 2.2

Multiply $- 1$ by $- 2$ .

$\frac{\log_{4} (81)}{\log_{4.24} (18) + 2}$

Step 3

Simplify the numerator.

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Step 3.1

Rewrite $\log_{4} (81)$ using the change of base formula.

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Step 3.1.1

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 $0$ .

$\frac{\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)}}{\log_{4.24} (18) + 2}$

Step 3.1.2

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

$\frac{\frac{\log (81)}{\log (4)}}{\log_{4.24} (18) + 2}$

Step 3.2

Rewrite $\log (81)$ as $\log (3^{4})$ .

$\frac{\frac{\log (3^{4})}{\log (4)}}{\log_{4.24} (18) + 2}$

Step 3.3

Expand $\log (3^{4})$ by moving $4$ outside the logarithm.

$\frac{\frac{4 \log (3)}{\log (4)}}{\log_{4.24} (18) + 2}$

Step 3.4

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

$\frac{\frac{4 \log (3)}{\log (2^{2})}}{\log_{4.24} (18) + 2}$

Step 3.5

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

$\frac{\frac{4 \log (3)}{2 \log (2)}}{\log_{4.24} (18) + 2}$

Step 3.6

Cancel the common factor of $4$ and $2$ .

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Step 3.6.1

Factor $2$ out of $4 \log (3)$ .

$\frac{\frac{2 (2 \log (3))}{2 \log (2)}}{\log_{4.24} (18) + 2}$

Step 3.6.2

Cancel the common factors.

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Step 3.6.2.1

Factor $2$ out of $2 \log (2)$ .

$\frac{\frac{2 (2 \log (3))}{2 (\log (2))}}{\log_{4.24} (18) + 2}$

Step 3.6.2.2

Cancel the common factor.

$\frac{\frac{2 (2 \log (3))}{2 \log (2)}}{\log_{4.24} (18) + 2}$

Step 3.6.2.3

Rewrite the expression.

$\frac{\frac{2 \log (3)}{\log (2)}}{\log_{4.24} (18) + 2}$

Step 3.7

Simplify $2 \log (3)$ by moving $2$ inside the logarithm.

$\frac{\frac{\log (3^{2})}{\log (2)}}{\log_{4.24} (18) + 2}$

Step 3.8

Raise $3$ to the power of $2$ .

$\frac{\frac{\log (9)}{\log (2)}}{\log_{4.24} (18) + 2}$

Step 4

Simplify the denominator.

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Step 4.1

Rewrite $\log_{4.24} (18)$ using the change of base formula.

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Step 4.1.1

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 $0$ .

$\frac{\frac{\log (9)}{\log (2)}}{\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} + 2}$

Step 4.1.2

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

$\frac{\frac{\log (9)}{\log (2)}}{\frac{\log (18)}{\log (4.24)} + 2}$

Step 4.2

Log base $10$ of $4.24$ is approximately $0.62736585$ .

$\frac{\frac{\log (9)}{\log (2)}}{\frac{\log (18)}{0.62736585} + 2}$

Step 4.3

Log base $10$ of $18$ is approximately $1.2552725$ .

$\frac{\frac{\log (9)}{\log (2)}}{\frac{1.2552725}{0.62736585} + 2}$

Step 4.4

Divide $1.2552725$ by $0.62736585$ .

$\frac{\frac{\log (9)}{\log (2)}}{2.000862 + 2}$

Step 4.5

Add $2.000862$ and $2$ .

$\frac{\frac{\log (9)}{\log (2)}}{4.000862}$

Step 5

Multiply the numerator by the reciprocal of the denominator.

$\frac{\log (9)}{\log (2)} \cdot \frac{1}{4.000862}$

Step 6

Divide $1$ by $4.000862$ .

$\frac{\log (9)}{\log (2)} \cdot 0.24994613$

Step 7

Multiply $\frac{\log (9)}{\log (2)} \cdot 0.24994613$ .

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Step 7.1

Combine $\frac{\log (9)}{\log (2)}$ and $0.24994613$ .

$\frac{\log (9) \cdot 0.24994613}{\log (2)}$

Step 7.2

Reorder $\log (9)$ and $0.24994613$ .

$\frac{0.24994613 \cdot \log (9)}{\log (2)}$

Step 7.3

Simplify $0.24994613 \cdot \log (9)$ by moving $0.24994613$ inside the logarithm.

$\frac{\log (9^{0.24994613})}{\log (2)}$

Step 8

Simplify the numerator.

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Step 8.1

Raise $9$ to the power of $0.24994613$ .

$\frac{\log (1.73184583)}{\log (2)}$

Step 8.2

Log base $10$ of $1.73184583$ is approximately $0.23850922$ .

$\frac{0.23850922}{\log (2)}$

Step 9

Log base $10$ of $2$ is approximately $0.30102999$ .

$\frac{0.23850922}{0.30102999}$

Step 10

Divide $0.23850922$ by $0.30102999$ .

$0.7923105$