Answer
$x=\dfrac{\ln3}{2}\approx0.549306$
Work Step by Step
$e^{4x}+4e^{2x}-21=0$
Rewrite the first term of this equation as $(e^{2x})^{2}$
$(e^{2x})^{2}+4e^{2x}-21=0$
Factor the equation:
$(e^{2x}+7)(e^{2x}-3)=0$
Set both factors equal to $0$ and solve each individual equation:
$e^{2x}+7=0$
Take the $7$ to the right side:
$e^{2x}=-7$
Since no values of $x$ make this first equation true, it has no solution. Let's move on to the second one:
$e^{2x}-3=0$
Take the $-3$ to the right side
$e^{2x}=3$
Apply $\ln$ to both sides:
$\ln e^{2x}=\ln3$
Take the exponent $2x$ to multiply in front of the $\ln$:
$2x\ln e=\ln3$
Since $\ln e=1$, this equation becomes:
$2x=\ln3$
Solve for $x$:
$x=\dfrac{\ln3}{2}\approx0.549306$
