What is the inverse of $y=\log(x+\sqrt{x^2+1})$?

Question

I'm unsure if I've done the first steps correctly.

If so, how would you proceed after getting $10^x = y+\sqrt{y^2+1}$?

Answer 1

Assuming the base $10$ logarithm is meant, and you have $10^x = y + \sqrt{y^2 + 1}$, that square root is the main source of difficulty.

The usual course of action is to get the square root by itself, and square, to get a nice equation that's polynomial in $y$:

\begin{align*} 10^x - y &= \sqrt{y^2 + 1}\\ (10^x)^2 - 2y10^x + y^2 &= y^2 + 1 \end{align*}

Now your $y^2$s can be cancelled from both sides, leaving the nice linear equation $10^{2x} - 2y10^x = 1$, and you shouldn't have too much trouble getting $y$ by itself, although it might look a bit funny.

---

Sidenote: If the logarithm were base $e$, this is the inverse of something called the hyperbolic sine function, $\sinh(x) = \dfrac{e^x - e^{-x}}{2}$. Yours should look similar, but with $10$ instead of $e$. You'll have to do a bit of cleanup to put it in the same format.