Precalculus Examples

$10^{\log_{p} (1)} + e^{2 \ln (3)} - \log_{9} (27)$

Step 1

Simplify each term.

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Step 1.1

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

$10^{0} + e^{2 \ln (3)} - \log_{9} (27)$

Step 1.2

Anything raised to $0$ is $1$ .

$1 + e^{2 \ln (3)} - \log_{9} (27)$

Step 1.3

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

$1 + e^{\ln (3^{2})} - \log_{9} (27)$

Step 1.4

Exponentiation and log are inverse functions.

$1 + 3^{2} - \log_{9} (27)$

Step 1.5

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

$1 + 9 - \log_{9} (27)$

Step 1.6

Logarithm base $9$ of $27$ is $\frac{3}{2}$ .

$1 + 9 - \frac{3}{2}$

Step 2

Find the common denominator.

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Step 2.1

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

$\frac{1}{1} + 9 - \frac{3}{2}$

Step 2.2

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

$\frac{1}{1} \cdot \frac{2}{2} + 9 - \frac{3}{2}$

Step 2.3

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

$\frac{2}{2} + 9 - \frac{3}{2}$

Step 2.4

Write $9$ as a fraction with denominator $1$ .

$\frac{2}{2} + \frac{9}{1} - \frac{3}{2}$

Step 2.5

Multiply $\frac{9}{1}$ by $\frac{2}{2}$ .

$\frac{2}{2} + \frac{9}{1} \cdot \frac{2}{2} - \frac{3}{2}$

Step 2.6

Multiply $\frac{9}{1}$ by $\frac{2}{2}$ .

$\frac{2}{2} + \frac{9 \cdot 2}{2} - \frac{3}{2}$

Step 3

Combine the numerators over the common denominator.

$\frac{2 + 9 \cdot 2 - 3}{2}$

Step 4

Simplify the expression.

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Step 4.1

Multiply $9$ by $2$ .

$\frac{2 + 18 - 3}{2}$

Step 4.2

Add $2$ and $18$ .

$\frac{20 - 3}{2}$

Step 4.3

Subtract $3$ from $20$ .

$\frac{17}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{17}{2}$

Decimal Form:

$8.5$

Mixed Number Form:

$8 \frac{1}{2}$