Algebra Examples

$\log_{5} (0.2) + \log_{0.5} (4)$

Step 1

Log base $5$ of $0.2$ is approximately $- 1$ .

$- 1 + \log_{0.5} (4)$

Step 2

Simplify each term.

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Step 2.1

Rewrite $\log_{0.5} (4)$ using the change of base formula.

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Step 2.1.1

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 $0$ .

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

Step 2.1.2

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

$- 1 + \frac{\log (4)}{\log (0.5)}$

Step 2.2

Log base $10$ of $0.5$ is approximately $- 0.30102999$ .

$- 1 + \frac{\log (4)}{- 0.30102999}$

Step 2.3

Move the negative in front of the fraction.

$- 1 - \frac{\log (4)}{0.30102999}$

Step 2.4

Log base $10$ of $4$ is approximately $0.60205999$ .

$- 1 - \frac{0.60205999}{0.30102999}$

Step 2.5

Divide $0.60205999$ by $0.30102999$ .

$- 1 - 1 \cdot 2$

Step 2.6

Multiply $- 1$ by $2$ .

$- 1 - 2$

Step 3

Subtract $2$ from $- 1$ .

$- 3$