Question 934844
{{{f(x) = log(3, (x + 1))}}}},  find inverse

since {{{f(x)=y}}}, we have

{{{y = log(3, (x + 1))}}},...swap {{{x}}} and {{{y}}}

{{{x= log(3, (y + 1))}}}...solve for {{{y}}}, change the base

{{{x= log( (y + 1))/log(3)}}}

{{{x*log(3)= log( (y + 1))}}}

{{{log(3^x)= log( (y + 1))}}} ...if log same, then

{{{3^x= y + 1}}}

{{{3^x-1= y }}}

so, inverse is:

{{{f^(-1) (x) = 3^x-1}}}

now find {{{f^(-1) (2) }}}

{{{f^(-1) (2) = 3^2-1}}}

{{{f^(-1) (2) = 9-1}}}

{{{f^(-1) (2) = 8}}}


{{{drawing( 600, 600, -10, 10, -10, 10,circle(2,8,.13),locate(2,8,p(2,8)), graph( 600, 600, -10, 10, -10, 10, log(3,(x + 1)), 3^x-1,3^x-1)) }}}