Question 941697
to find the inverse of {{{y=  log(2,8^(x-1))}}}, first swap {{{x}}} and {{{y}}}

{{{x=  log(2,8^(y-1))}}}...now solve for {{{y}}}

{{{x=  log(2,8^(y-1))}}}...change to base {{{10}}}

{{{x=  log(8^(y-1))/log(2)}}}

{{{x*log(2)=  log(8^(y-1))}}}

{{{log(2^x)=  log(8^(y-1))}}}...since log same, we have

{{{2^x= 8^(y-1)}}}....we can write {{{8}}} as {{{2^3}}}

{{{2^x= (2^3)^(y-1)}}}

{{{2^x= 2^(3(y-1))}}}.....if bases same then exponents are same too

{{{x= 3(y-1)}}}

{{{x= 3y-3}}}

{{{x+3= 3y}}}

{{{y=x/3+1}}}..=>the inverse of {{{y=  log(2,8^(x-1))}}}