Question 1208782
Assumed, that symbol really should be variable, x.

{{{log(((2x+1)/(3x-2)))=1}}}


You want base 10?


{{{10^1=(2x+1)/(3x-2)}}}


{{{10(3x-2)=2x+1}}}


{{{30x-20=2x+1}}}


{{{28x=21}}}

{{{4x=3}}}

{{{x=3/4}}} but you should check if this makes sense in your original equation.