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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!

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    $\begingroup$ Both sides times e^x. Could you take it from here? $\endgroup$ – Vim May 11 '15 at 4:06
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    $\begingroup$ suppose $y=e^x$. $\endgroup$ – John Joy May 11 '15 at 13:19
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$$e^x-e^{-x}=\frac{3}{2}$$ 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$.

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  • $\begingroup$ 2nd line, I assume you mean $e^{2x}$ not $e^{x^2}$ $\endgroup$ – uqtredd1 May 11 '15 at 4:14
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    $\begingroup$ In the last line you forgot to multiply the -1 by two. $\endgroup$ – Roman Reiner May 11 '15 at 6:37
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Hint: Mutiply by $e^x$ and rearrange to get a quadratic equation in $e^x$.

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

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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!

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