Question

log (base 9) 1/81

Answer 1

Remember that \[\log_a x = y\] and \[x = a^y\]
See that you assign your numbers to the right variables!
a = 9
x = 1/81

If we rearrange this a little, we get \[\frac{ 1 }{ 81 } = 9^y\]
Now it's just a matter of figuring out what to raise 9 to so we get out 1/81.

Consider that \[9^2 = 81\] Knowing that, we can apply what we know about negative exponents: \[x^{-n} = \frac{ 1 }{ x^n }\] And from there, safely determine that \[9^{-2} = \frac{ 1 }{ 81 }\]So at the end of it all,
\[\log_9 \frac{ 1 }{ 81 } = -2\]

Hope this helps!