Question 369701
{{{e^x-e^(-x)=-4}}}
{{{e^(2x)-1=-4e^x}}}
{{{e^(2x)+4e^x-1=0}}}
{{{u^2+4u-1=0}}} where {{{u=e^x}}}
{{{u^2+4u+4-1-4=0}}}
{{{(u+2)^2=5}}}
{{{u+2=0 +- sqrt(5)}}}
{{{u=-2 +- sqrt(5)}}}
{{{e^x=-2 +- sqrt(5)}}}
{{{highlight(x=ln(-2+sqrt(5)))}}}
The negative solution was removed since the natural log function requires a positive argument.