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. 
$$
\frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{3}{4}
$$
 A: 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.
A: We know that $$\sinh x =\frac {e^x -e^{-x}}{2} $$ and that  $$\cosh x =\frac {e^x + e^{-x}}{2} $$ Can you now proceed where $\sinh x $ and $\cosh x $ are hyperbolic trigonometric functions?

We can otherwise put $k =e^x $ and then get our equation as $$\frac {k-\frac {1}{k}}{k+\frac {1}{k}} =\frac {3}{4} $$ $$4 (k^2-1) =3 (k^2+1) $$ $$\Rightarrow k = \sqrt {7} \text { as an acceptable solution} $$ Hope it helps. 
A: Here is a solution without the use of hyperbolic trigonometric functions.
We have:
$$\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{3}{4}$$
Let's multiply both the numerator and denominator of the left hand side by $e^x$:
$$\frac{(e^x-e^{-x})\cdot e^x}{(e^x+e^{-x})\cdot e^x}=\frac{3}{4}$$
$$\frac{e^{2x}-1}{e^{2x}+1}=\frac{3}{4}$$
Now, you can substitute $u=e^{2x}$, and obtain a linear equation after doing cross-multiplication:
$$\frac{u-1}{u+1}=\frac{3}{4}$$
You can now easily solve for $u$, and substitute back $u=e^{2x}$. Then, solve for $x$.
If you have any questions, please do not hesitate to ask.
