Question 888937
{{{16^x+4^(x+1)-3=0}}}
{{{(4^2)^x+4^(x+1)-3=0}}}
{{{4^(2x)+4*4^(x)-3=0}}}
Substitute, let {{{u=4^x}}}
{{{u^2+4u-3=0}}}
{{{(u^2+4u+4)-3-4=0}}}
{{{(u+2)^2=7}}}
{{{u+2=0 +- sqrt(7)}}}
{{{u=-2 +- sqrt(7)}}}
{{{4^x=-2 +- sqrt(7)}}}
Only the positive solution makes sense for this problem.
{{{4^x=sqrt(7)-2}}}
{{{highlight(x=log((sqrt(7)-2))/log((4)))}}}