Question 48096
{{{log(2x,16) = -2}}}
{{{16 = (2x)^(-2)}}}
{{{16 = 1/(4x^2)}}}
{{{1/16 = 4x^2}}}
{{{1/64 = x^2}}}
+-{{{1/8 = x}}}
'b' can not be negative so your solution is: {{{1/8 = x}}}