Precalculus Examples

$\log_{3} (\frac{4}{3})$

Step 1

Rewrite $\frac{4}{3}$ as $4 \cdot 3^{- 1}$ .

$\log_{3} (4 \cdot 3^{- 1})$

Step 2

Rewrite $\log_{3} (4 \cdot 3^{- 1})$ as $\log_{3} (4) + \log_{3} (3^{- 1})$ .

$\log_{3} (4) + \log_{3} (3^{- 1})$

Step 3

Use logarithm rules to move $- 1$ out of the exponent.

$\log_{3} (4) - \log_{3} (3)$

Step 4

Logarithm base $3$ of $3$ is $1$ .

$\log_{3} (4) - 1 \cdot 1$

Step 5

Multiply $- 1$ by $1$ .

$\log_{3} (4) - 1$

Step 6

Simplify each term.

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Step 6.1

Rewrite $\log_{3} (4)$ as $\log_{3} (2^{2})$ .

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

Step 6.2

Expand $\log_{3} (2^{2})$ by moving $2$ outside the logarithm.

$2 \log_{3} (2) - 1$

Step 7

Simplify each term.

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Step 7.1

Rewrite $\log_{3} (2)$ using the change of base formula.

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Step 7.1.1

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 $0$ .

$2 \log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} - 1$

Step 7.1.2

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

$2 \frac{\log (2)}{\log (3)} - 1$

Step 7.2

Multiply $2 \frac{\log (2)}{\log (3)}$ .

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Step 7.2.1

Combine $2$ and $\frac{\log (2)}{\log (3)}$ .

$\frac{2 \log (2)}{\log (3)} - 1$

Step 7.2.2

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

$\frac{\log (2^{2})}{\log (3)} - 1$

Step 7.3

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

$\frac{\log (4)}{\log (3)} - 1$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\frac{\log (4)}{\log (3)} - 1$

Decimal Form:

$0.26185950 \dots$