Question 735036
For simplicity of typing I am leaving the base off the logs.

Use the log rules:

first: {{{rloga=log(a^r)}}}

so {{{(1/2)log 25=log(25^(1/2))=log(sqrt(25))=log5}}}
and {{{5log2=log(2^5)=log32}}}

Now use the rule: {{{loga-logb=log(a/b)}}}


{{{log5-log32=log(5/32)}}}

so {{{logx=(1/2)log 25 - 5log2 = log(25^(1/2))-log(2^5)=log5-log32=log(5/32)}}}

so {{{logx=log(5/32)}}}

and {{{x=5/32}}}