Question

MAJORRR HELP !!!!!

Answer 1

$\frac{3}{4} = log_{2}(\sqrt[4]{8} )\\-4 = log_{3} \frac{1}{81} \\-6= -3log_{5} 25\\\frac{1}{3} = log_{6} (\sqrt[3]{6} )$

Here's how you solve it!

$log_{2} \sqrt[4]{8}$

Write it in exponential form

$log_{2} (2 \frac{3}{4} )$

Then simplify

$\frac{3}{4}$

$log_{3} \frac{1}{81}$

Write in exponential form

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

Simplify

-4

$-3log_{5} 25$

Write in exponential form

$-3log_{5} (5^{2} )$

Simplify

-3 * 2 = -6

-6

$log_{6} \sqrt[3]{6}$

Write in exponential form

$log_{6} (6\frac{1}{3} )$

Simplify

$\frac{1}{3}$

Hope this helps! :3