Question

how do you solve log base 3 of 1/9 = x ?
i don't know how to do fractions :/ so imma be asking a lot cos i really don't get it

Answer 1

exactly the same way we did
\[\log_2(\frac{1}{2})=-1\]

Answer 2

start with equivalent exponential form
\[3^x=\frac{1}{9}\]

Answer 3

then think of what power you would raise 3 to to get the answer of 1/9

Answer 4

\[3^2=9\] and so
\[3^{-2}=\frac{1}{9}\] and so your answer is
\[\log_3(\frac{1}{9})=-2\]

Answer 5

what if it's the opposite way now ? like , log base 9 of 1/3 = x ?