Precalculus Examples

$\frac{\ln (e^{2}) - \log_{3} (81)}{\ln (e^{x^{2}}) - \log (10)}$

Step 1

Simplify the numerator.

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Step 1.1

Use logarithm rules to move $2$ out of the exponent.

$\frac{2 \ln (e) - \log_{3} (81)}{\ln (e^{x^{2}}) - \log (10)}$

Step 1.2

The natural logarithm of $e$ is $1$ .

$\frac{2 \cdot 1 - \log_{3} (81)}{\ln (e^{x^{2}}) - \log (10)}$

Step 1.3

Multiply $2$ by $1$ .

$\frac{2 - \log_{3} (81)}{\ln (e^{x^{2}}) - \log (10)}$

Step 1.4

Logarithm base $3$ of $81$ is $4$ .

$\frac{2 - 1 \cdot 4}{\ln (e^{x^{2}}) - \log (10)}$

Step 1.5

Multiply $- 1$ by $4$ .

$\frac{2 - 4}{\ln (e^{x^{2}}) - \log (10)}$

Step 1.6

Subtract $4$ from $2$ .

$\frac{- 2}{\ln (e^{x^{2}}) - \log (10)}$

Step 2

Simplify the denominator.

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Step 2.1

Use logarithm rules to move $x^{2}$ out of the exponent.

$\frac{- 2}{x^{2} \ln (e) - \log (10)}$

Step 2.2

The natural logarithm of $e$ is $1$ .

$\frac{- 2}{x^{2} \cdot 1 - \log (10)}$

Step 2.3

Multiply $x^{2}$ by $1$ .

$\frac{- 2}{x^{2} - \log (10)}$

Step 2.4

Logarithm base $10$ of $10$ is $1$ .

$\frac{- 2}{x^{2} - 1 \cdot 1}$

Step 2.5

Multiply $- 1$ by $1$ .

$\frac{- 2}{x^{2} - 1}$

Step 2.6

Rewrite $x^{2} - 1$ in a factored form.

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Step 2.6.1

Rewrite $1$ as $1^{2}$ .

$\frac{- 2}{x^{2} - 1^{2}}$

Step 2.6.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$\frac{- 2}{(x + 1) (x - 1)}$

Step 3

Move the negative in front of the fraction.

$- \frac{2}{(x + 1) (x - 1)}$