How did I lose some solutions?

Question

Solve the following equation for x. $$\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}=y$$

My solution ($u=e^x$): $$\frac{e^{x}+\frac{1}{e^{x}}}{e^{x}-\frac{1}{e^{x}}}=y\\ \frac{u+\frac{1}{u}}{u-\frac{1}{u}}.\frac{u}{u}=y\\ \frac{u^2+1}{u^2-1} = y\\ u^2+1=y(u^2-1)\\ u^2-yu^2=-y-1\\ u^2=\frac{y+1}{y-1}\\ u=\pm\sqrt{\frac{y+1}{y-1}}\\ u=e^x=\sqrt{\frac{y+1}{y-1}}\\ x=\ln(\frac{y+1}{y-1})^{\frac{1}{2}}\\ x=\frac{1}{2}(\ln(y+1)-\ln(y-1))$$

In my solution y must be greater than 1, but in the original equation above, y can be anything except zero. How did I lose the rest of the solutions?

Answer 1

No solutions exist if $-1\leq y \leq 1$. To see this, compare the values of the denominator and numerator in the LHS of the first equation e.g. in separate cases $x<0$ and $x>0$.

Now an interesting question is why did you miss the case $y<-1$? Note that you used $\ln(a/b)= \ln a - \ln b$. However, this is only true when $a$ and $b$ are nonnegative. You have to treat the case $a,b<0$ separately. For example, $\ln \left(\frac{-2}{-2}\right)$ is defined (it's just $\ln 1$), but $\ln(-2)$ is not!