Evaluate log base 2 of 1/64+ log base 2 of 2 square root of 2- log base 3 of 1/81

$\log_{2} (\frac{1}{64}) + \log_{2} (2 \sqrt{2}) - \log_{3} (\frac{1}{81})$

Step 1

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

$\log_{2} (\frac{1}{64} (2 \sqrt{2})) - \log_{3} (\frac{1}{81})$

Step 2

Simplify each term.

Cancel the common factor of $2$ .

Factor $2$ out of $64$ .

$\log_{2} (\frac{1}{2 (32)} (2 \sqrt{2})) - \log_{3} (\frac{1}{81})$

Factor $2$ out of $2 \sqrt{2}$ .

$\log_{2} (\frac{1}{2 (32)} (2 (\sqrt{2}))) - \log_{3} (\frac{1}{81})$

Cancel the common factor.

$\log_{2} (\frac{1}{2 \cdot 32} (2 \sqrt{2})) - \log_{3} (\frac{1}{81})$

Rewrite the expression.

$\log_{2} (\frac{1}{32} \sqrt{2}) - \log_{3} (\frac{1}{81})$

Combine $\frac{1}{32}$ and $\sqrt{2}$ .

$\log_{2} (\frac{\sqrt{2}}{32}) - \log_{3} (\frac{1}{81})$

Logarithm base $3$ of $\frac{1}{81}$ is $- 4$ .

$\log_{2} (\frac{\sqrt{2}}{32}) - - 4$

Multiply $- 1$ by $- 4$ .

$\log_{2} (\frac{\sqrt{2}}{32}) + 4$

Step 3

Rewrite $\log_{2} (\frac{\sqrt{2}}{32})$ 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 $0$ .

$\log_{a} (x) = \frac{\log_{b} (x)}{\log_{b} (a)} + 4$

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

$\frac{\log (\frac{\sqrt{2}}{32})}{\log (2)} + 4$

Step 4

The result can be shown in multiple forms.

Exact Form:

$\frac{\log (\frac{\sqrt{2}}{32})}{\log (2)} + 4$

Decimal Form:

$- 0.5$