Question 917848
This is a quadratic equation in disguise.
Use a substitution,
{{{u=4^x}}}
{{{u^2=(4^x)^2=4^(2x)=16^x}}}
So then,
{{{u^2+u+1-3=0}}}
{{{u^2+u-2=0}}}
{{{(u+2)(u-1)=0}}}
Two "u" solutions:
{{{u+2=0}}}
{{{u=-2}}}
{{{4^x=-2}}}
No solutions in x here.
{{{u-1=0}}}
{{{u=1}}}
{{{4^x=1}}}
{{{highlight(x=0)}}}