Question 1186695


If {{{2^y=10}}}, find the value of{{{ 8^(2y-1)}}}

first find {{{y}}}

{{{2^y=10}}}....take log of both sides

{{{log(2^y)=log(10)}}}

{{{y*log(2)=1}}}

{{{y=1/log(2)}}}


substitute in

{{{8^(2y-1)=8^(2(1/log(2))-1)= 8^5.64385618977472=125000}}}