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My homework has tasked me with finding $x$ when $\cosh x=5/3$. I know that the solution is $\ln (3)$, but I can't figure out how to solve it myself. The furthest I can simplify it is the following:

$\frac{e^x+e^{-x}}{2} = 5/3$ $ e^x+e^{-x} = 10/3$ $e^x = \frac{10}{3}-e^{-x}$ $x = \ln \left(\frac{10}{3}-e^{-x} \right)$

Now, if I put this into Wolfram Alpha it tells me that the answer is $\ln(3)$, but it doesn't tell me how it solved that. Also, I'm guessing there may be another way to solve this by taking a different route than the above.

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    @MattMunson: Not quite: The original title was "How to show that $x=\ln(3/10 - e^{-x}) = \ln3$?" which is also not exactly asking how to show that *if* the first equality holds, *then* $x=\ln 3$.2011-11-12

3 Answers 3

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Start with $e^x+e^{-x}={10\over3}$

Multiply both sides by $e^x$:

$ e^x\cdot e^x+e^x\cdot e^{-x}={10\over 3}e^x $ Simplify: $ (e^x)^2+1={10\over 3}e^x. $

Let $u=e^x$, then $ u^2+1={10\over3}u $

or $ 3u^2-10u+3=0. $ This has solutions $u=3$ and $u=1/3$.

So $e^x=3$ or $e^x=1/3$.

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    @J.M. No, I don't know how to complete the square.2011-11-12
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Here is something a little different that gives us a chance to play with hyperbolic functions.

Recall that $\cosh x=\frac{e^x+e^{-x}}{2} \qquad \text{and}\qquad \sinh x=\frac{e^x-e^{-x}}{2}.$ The fact that $\cosh(-x)=\cosh x$ and $\sinh(-x)=-\sinh x$ can be verified directly from the definitions.

The result for $\cosh x$ shows that if we find a solution $x$ of the equation $\cosh x=a$, then $-x$ is also a solution of the equation.

From the definitions of $\cosh x$ and $\sinh x$, it is not hard to verify that $\cosh^2 x-\sinh^2x=1.\qquad(\ast)$ This identity, so reminiscent of $\cos^2 x+\sin^2 x=1$, is the key to the importance of $\cosh x$ and $\sinh x$, and to the computation that follows.

Enough of previews. It is time for the movie.


If $\cosh x =\frac{5}{3}$, then by $(\ast)$ $\sinh^2 x=\cosh^2 x-1=\frac{25}{9}-1=\frac{16}{9}$, and therefore $\sinh x=\pm \frac{4}{3}$. Thus $\frac{e^x+e^{-x}}{2}=\frac{5}{3} \qquad \text{and}\qquad \frac{e^x-e^{-x}}{2}=\pm \frac{4}{3}.$ Add. We get $e^x=\frac{5}{3}\pm \frac{4}{3}.$ Thus $e^x=\frac{9}{3}=3$ or $e^x=\frac{1}{3}$. It follows that $x=\ln 3$ or $x=\ln(1/3)=-\ln 3$.

Comment: Exactly the same method can be used to solve $\cosh x=a$. (There is no real solution if $a<1$.)

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as @David Mitra said you made a light misstep in the third step

$e^x+\frac1{e^x}=\frac {10}{3}$

$3e^{2x}+3={10} e^x$

$3({e^x})^2-{10} e^x+3=0$

then its just solving $e^x$ like any other quadratic (don't forget to eliminate the negative solution (if any))