Question 1131605

{{{(y-1)log(4) = ylog(16)}}}

{{{log(4^(y-1)) = log(16^y)}}}


{{{log(4^(y-1)) = log((4^2)^y)}}}


{{{log(4^(y-1)) = log(4^(2y))}}}........log same, then


{{{4^(y-1) = 4^(2y)}}}............bases same, then


{{{y-1 = 2y}}}


{{{-1 = 2y-y}}}


{{{y=-1 }}}