Solve for x in each equation. Question 1. $\displaystyle e^{5x} = 46.38^{(2.6)}$ Question 2. $\displaystyle e^x-e^{-x}=4$
Question 2:
Multiply through by $\displaystyle e^x$
$\displaystyle e^x-e^{-x}=4$
so:
$\displaystyle e^{2x}-1=4 e^x$
then let $\displaystyle y=e^x$ so:
$\displaystyle
y^2-4y-1=0
$
solving this gives $\displaystyle y=2 \pm \sqrt{5}$, but as $\displaystyle y>0$ only $\displaystyle y=2+\sqrt{5}$ is a valid solution if we want $\displaystyle x$ to be real. Then $\displaystyle x=\ln(2+\sqrt{5})$
RonL