Precalculus Examples

$\log_{3} (5) + 5 \log_{3} (2)$

Step 1

Simplify each term.

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Step 1.1

Simplify $5 \log_{3} (2)$ by moving $5$ inside the logarithm.

$\log_{3} (5) + \log_{3} (2^{5})$

Step 1.2

Raise $2$ to the power of $5$ .

$\log_{3} (5) + \log_{3} (32)$

Step 2

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

$\log_{3} (5 \cdot 32)$

Step 3

Multiply $5$ by $32$ .

$\log_{3} (160)$

Step 4

Rewrite $\log_{3} (160)$ 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 $0$ .

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

Step 4.2

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

$\frac{\log (160)}{\log (3)}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{\log (160)}{\log (3)}$

Decimal Form:

$4.61962228 \dots$