Question 964624
This is a quadratic equation in disguise.
Let {{{u=e^x}}}
{{{u(2u-1)=10}}}
{{{2u^2-u-10=0}}}
{{{(u+2)(2u-5)=0}}}
Two solutions:
{{{u+2=0}}}
{{{u=-2}}}
{{{e^x=-2}}}
Not possible.
{{{2u-5=0}}}
{{{2u=5}}}
{{{u=5/2}}}
{{{e^x=(5/2)}}}
{{{x=ln(5/2)}}}
{{{highlight(x=ln(5)-ln(2))}}}