Question

1/81 = b^(-4) I know the answer is 3, but I need to show the work

Answer 1

help please!!!!!!

Answer 2

will medal (if i can figure out how)

Answer 3

\[b^{-4} = \frac{ 1 }{ b^4 }\]
\[\frac{ 1 }{ 81 } =\frac{ 1 }{ b^4 }\]
\[81 = {b^4}\]
\[\sqrt{81} = \sqrt{b^2}\]
\[9 = b^2\]
\[\sqrt{9} = \sqrt{b}\]
b = 3

Answer 4

thank you thank you thank you!

Answer 5

anytime :)

Answer 6

@sangya21 okay, real quick, how did u get b^2 out of b^4 on line four?

Answer 7

\[81 = b^2 *b^2\]
\[\sqrt{81} =\sqrt{b^4}\]
\[9 ={b^2}\]

Answer 8

@sangy21 i see it now. u worked down the roots

Answer 9

Yups.