Evaluate log base 0.33 of 81

$\log_{0.33} (81)$

Step 1

Rewrite $\log_{0.33} (81)$ using the change of base formula.

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 $0$ .

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

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

$\frac{\log (81)}{\log (0.33)}$

Step 2

Log base $10$ of $0.33$ is approximately $- 0.48148606$ .

$\frac{\log (81)}{- 0.48148606}$

Step 3

Move the negative in front of the fraction.

$- \frac{\log (81)}{0.48148606}$

Step 4

Log base $10$ of $81$ is approximately $1.90848501$ .

$- \frac{1.90848501}{0.48148606}$

Step 5

Divide $1.90848501$ by $0.48148606$ .

$- 1 \cdot 3.96373888$

Step 6

Multiply $- 1$ by $3.96373888$ .

$- 3.96373888$