Question 842695
Use a substitution, {{{u=log(2,x)}}},
{{{u^2-2u-15=0}}}
{{{(u-5)(u+3)=0}}}
Two possible solutions:
{{{u-5=0}}}
{{{u=5}}}
{{{log(2,x)=5}}}
{{{x=2^5=32}}}
and
{{{u+3=0}}}
{{{u=-3}}}
{{{log(2,x)=-3}}}
{{{x=2^(-3)=1/8}}}