Precalculus Examples

$\log_{2} (16) - \log_{9} (3) + \ln (e^{7})$

Step 1

Simplify each term.

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Step 1.1

Logarithm base $2$ of $16$ is $4$ .

$4 - \log_{9} (3) + \ln (e^{7})$

Step 1.2

Logarithm base $9$ of $3$ is $\frac{1}{2}$ .

$4 - \frac{1}{2} + \ln (e^{7})$

Step 1.3

Use logarithm rules to move $7$ out of the exponent.

$4 - \frac{1}{2} + 7 \ln (e)$

Step 1.4

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

$4 - \frac{1}{2} + 7 \cdot 1$

Step 1.5

Multiply $7$ by $1$ .

$4 - \frac{1}{2} + 7$

Step 2

Find the common denominator.

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Step 2.1

Write $4$ as a fraction with denominator $1$ .

$\frac{4}{1} - \frac{1}{2} + 7$

Step 2.2

Multiply $\frac{4}{1}$ by $\frac{2}{2}$ .

$\frac{4}{1} \cdot \frac{2}{2} - \frac{1}{2} + 7$

Step 2.3

Multiply $\frac{4}{1}$ by $\frac{2}{2}$ .

$\frac{4 \cdot 2}{2} - \frac{1}{2} + 7$

Step 2.4

Write $7$ as a fraction with denominator $1$ .

$\frac{4 \cdot 2}{2} - \frac{1}{2} + \frac{7}{1}$

Step 2.5

Multiply $\frac{7}{1}$ by $\frac{2}{2}$ .

$\frac{4 \cdot 2}{2} - \frac{1}{2} + \frac{7}{1} \cdot \frac{2}{2}$

Step 2.6

Multiply $\frac{7}{1}$ by $\frac{2}{2}$ .

$\frac{4 \cdot 2}{2} - \frac{1}{2} + \frac{7 \cdot 2}{2}$

Step 3

Combine the numerators over the common denominator.

$\frac{4 \cdot 2 - 1 + 7 \cdot 2}{2}$

Step 4

Simplify each term.

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Step 4.1

Multiply $4$ by $2$ .

$\frac{8 - 1 + 7 \cdot 2}{2}$

Step 4.2

Multiply $7$ by $2$ .

$\frac{8 - 1 + 14}{2}$

Step 5

Simplify by adding and subtracting.

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Step 5.1

Subtract $1$ from $8$ .

$\frac{7 + 14}{2}$

Step 5.2

Add $7$ and $14$ .

$\frac{21}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$\frac{21}{2}$

Decimal Form:

$10.5$

Mixed Number Form:

$10 \frac{1}{2}$