Find $x$ if $\sinh(x)=2$

Question

I want to know how to find $x$ if $\sinh(x)=2$. I already know that $\sinh(x) = \dfrac{e^x -e^{-x}}{2}$. Hence, $$\frac{e^x -e^{-x}}{2} = 2 \implies e^x -e^{-x}=4$$ but I don't know what should I do then.

Answer 1

The answer is $x = \sinh^{-1}(2)$. Note that it can also be written as follows: We have $e^x - e^{-x} = 4$. Let $y=e^x$. We then have $$y-\dfrac1y = 4 \implies y^2-1 = 4y \implies y^2-4y-1 = 0 \implies (y-2)^2=5 \implies y=2+\sqrt5$$ (We discard the root $y=2-\sqrt5$ assuming that we are after a real $x$ and hence $e^x > 0$). Hence, we have $$e^x = 2 + \sqrt5 \implies x = \ln(2+\sqrt5)$$