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Using the definition $\sinh x = \dfrac{e^x-e^{-x}}{2},\;$ let's say we want to solve $\;y = \sinh x \;$ for $x$.

It's not hard to show that $\;\sinh x \;$ is bijective, so this should have exactly one solution:

$$\begin{align} y &= \frac{e^x-e^{-x}}{2} \\ \\ y &= e^x - e^{-x} \\ \\ 2ye^x &= (e^x)^2 - 1 \\ \end{align}$$ $$(e^x)^2 - 2y e^x -1 = 0$$ $$ e^x = y \pm \sqrt{1+y^2} $$

Now, the solution with "$-$" cannot be, since it's always negative and we can't take the log. Therefore, we get that $y = \sinh x \iff x = \ln(y+\sqrt{1+y^2})$.

What I don't understand is, why are there two solutions to the quadratic? You would think that since we start with the equation for a bijective function and apply reversible operations to it (multiplication by $2$, multiplication by $e^x$, subtracting $2ye^x$) we would get something that has only one solution, but somehow we get two and one has to be discarded. Why does that happen?

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You get one complex solution – Amr Dec 26 '12 at 21:57
And, if you study the logarithm further, you get infinitely many complex solutions... – GEdgar Dec 26 '12 at 22:36

You only get two solutions because you use the general method of solving quadratic equations over the all of the reals to solve $(e^x)^2 - 2y e^x -1 = 0$, after substituting $z$ for $e^x$. But $z$ does not range over all of the reals, so you should really either use a specific method to solve a quadratic equations over the positive reals, or use an ad hoc method to solve this equation without using any substitution. Of course a method to solve a quadratic equations over the positive reals will basically just consist of using those values of the quadratic formula that are positive. But one could state the general fact that with all variables ranging over the positive reals the equation $z^2=2bz+c$ has a unique solution $z=b+\sqrt{b^2+c}$, and using that fact with $z=e^x$ one would never see a second solution. What I want to say is: the spurious solution is more an artifact of the method used to find the solution than a characteristic of the original equation itself.

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I have just posted the opposite opinion. That the "spurious" solutions are actually valid complex solutions should count in favor of the method; it gives us more correct material than we put in. – ASCII Advocate Aug 13 '15 at 22:34

Here is a diametrically opposite interpretation from that of the first (and until now the only) answer.

The method used is correct; solving the equation one shows that $e^x$ has one of two possible real values. If a solution exists, then it is in such-and-such subset of $\mathbb{R}$. That is a correct deduction.

This correct application of the quadratic equation is insufficient to isolate a unique $x$. This may be surprising because $\sinh$ is injective on the reals. But $\sinh$ is not uniquely invertible on the complex numbers, or matrices, other domains extending the reals but satisfying the same algebraic laws used in solving the quadratic equation. Thus, the assumption that $x$ is real has to also be used as a separate piece of argument, and that means $e^x$ is positive. This pins down a single value of $x$.

The algebra does not "know" that $x$ was restricted to $\mathbb{R}$, but that is not a problem with the algebra.

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