Question 1125603

{{{ f(x)=log(2 ,(10x-20)) }}}

change the base, replace {{{f(x) }}}with {{{y}}}:


{{{ y=log((10x-20))/ log((2))}}}

Interchange the variables:

{{{ x=log((10y-20))/ log((2))}}}

solve or {{{y}}}

{{{ x*log((2))=log((10y-20))}}}

{{{ log((2^x))=log((10y-20))}}} ...since log equals, we will have

{{{ 2^x=10y-20}}} 

.{{{ 2^x+20=10y}}} 

{{{ y=(2^x+20)/10}}} 

{{{ y=(1/10)(2^x+20)}}}->your inverse 

so, replace {{{y}}} with {{{ f^-1(x)}}}:

{{{ f^-1(x)=(1/10 )(2^x + 20)}}}