Question 389223
{{{4(4^x)+127(2^x)=32}}}
Use a substitution,
Let {{{u=2^x}}}, then {{{u^2=(2^x)^2=2^(2x)=4^x}}}
{{{4u^2+127u=32}}}
{{{4u^2+127u-32=0}}}
{{{(4u-1)(u+32)=0}}}
Two solutions in {{{u}}}:
{{{4u-1=0}}}
{{{4u=1}}}
{{{u=1/4}}}
{{{2^x=1/4=1/2^2=2^(-2)}}}
{{{highlight(x=-2)}}}
.
.
{{{u+32=0}}}
{{{u=-32}}}
{{{2^x=-32}}}
No solution.