Question 1180696

{{{2^(x-2)= 32}}}

{{{b^y = x}}} is equivalent to {{{y = log(b,x) }}}

in your case {{{b=2}}}, {{{y=(x-2)}}}, and {{{x=32}}}

{{{x-2= log(2,32) }}}

{{{x= log(2,32) +2}}}-> log form

solution:

{{{x= log(2,2^5) +2}}}
{{{x= 5log(2,2) +2}}}
{{{x= 5*1 +2}}}
 {{{x= 7 }}}