Find $x$ in equation having Euler's Number I have to find $x$ in this equation. I have already tried cross multiplying but I am not sure what to do after. 
$$
\frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{3}{4}
$$

Let $z=e^x$. Then
$$\frac{z-z^{-1}}{z+z^{-1}}=\frac34,$$
which leads to the quadratic equation
$$4(z^2-1)=3(z^2+1),$$
or 
$$z^2=7.$$
Hence $x=\dfrac{\log7}2$, which is the only solution.
