Solving an exponential equation involving e: $e^x-e^{-x}=\frac{3}{2}$

Question

In a previous exam, my professor had the question

\begin{equation*} e^x-e^{-x}=\frac{3}{2}. \end{equation*}

I attempted to take the natural log of both side to solve it, but evidently that was incorrect... how does one start to go about solving this type of problem? Any help or advice would be greatly appreciated. Thanks!

Answer 1

$$e^x-e^{-x}=\frac{3}{2}$$ Multiply by $e^x$. $${\left(e^{x}\right)^2}-1=\frac{3e^x}{2}$$ Let $u=e^x$ $$u^2-\frac{3u}{2}-1=0$$ $$2u^2-3u-2=0$$ Now solve for $u$, and back substitute into $u=e^x$. Consider only positive solutions for $ u$.

Answer 2

Hint: Mutiply by $e^x$ and rearrange to get a quadratic equation in $e^x$.

Answer 3

This is the same as solving $$2\sinh x=\frac 32$$ so you could use the logarithmic formula for $\operatorname{arsinh} x$.

Answer 4

Your equation can be re-written as

$e^x-\frac{1}{e^x}=\frac{3}{2}$

$e^{2x}-1=\frac{3}{2}\times e^x$

$2e^{2x}-2=3e^x$

Now consider,

$e^x=y$

then,

$2y^2-3y-2=0$

$2y^2-4y+y-2=0$

$2y(y-2)+1(y-2)=0$

Giving you,

$y=-\frac{1}{2}$ or $y=2$

That makes,

$e^x=2$

Voila!