Question 743185
log functions are inverses of exponential functions. We have two choices. In this case we'll use log base 8 (can you see why?).

{{{(5/9)^x=8^(1-x)}}}
{{{log_base 8((5/9)^x)=log_base 8((8)^(1-x)))}}}

First we rewrite so that out exponents are now products and we use the fact that
 
{{{log_base 8((8))=1}}}

to simplify our equation and we get:

{{{x*log_base 8(5/9)=1-x}}}
{{{x*log_base 8(5/9)+x=1}}}
{{{x*(log_base 8(5/9)+1)=1}}}
{{{x*(log_base 8(5/9)+1)/(log_base 8(5/9)+1)=1/(log_base 8(5/9)+1)}}}
{{{x=1/(log_base 8(5/9)+1)}}}