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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.

Eric Naslund
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Matt Munson
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    Multiply your first equation by $e^x$ and you have a quadratic in terms of $e^x$. – David Mitra Nov 11 '11 at 21:54
  • @David. I don't see how, but then I'm not very familiar with quadratics – Matt Munson Nov 11 '11 at 21:58
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    You don't actually want to show that $\ln(10/3 - e^{-x})$ is $\ln 3$ (it can't be; the left hand side is a nonconstant function, the right hand side a constant); you want to show that the only solution to $x=\ln(10/3 - e^{-x})$ is $x=\ln 3$, which is a different statement. (Compare: showing $x=2x$ is only true when $x=0$ or $x=1$ is not the same thing as showing that $2x=0$ or $2x=1$). – Arturo Magidin Nov 11 '11 at 22:57
  • @Arturo That was in the original title that others modified.But noted nonetheless. – Matt Munson Nov 12 '11 at 01:01
  • @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 $x=\ln 3$. – Arturo Magidin Nov 12 '11 at 02:22

3 Answers3

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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 $e^x=1/3$.

David Mitra
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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 $a<1$.)

Quixotic
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André Nicolas
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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))

Michael Hardy
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ratchet freak
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