# How do I solve $x=\log^e{(x+1)}$ analytically?

How do I solve the following, analytically?

$$x=\log^e{(x+1)}$$

It looks like it should be simple, but whether I take the $e$th root of each side or take the $\log$ of each side (ending up with a $\log\log$), I get stuck—and both approaches seem naïve, anyhow.

It appears to have two solutions: In case it raises some eyebrows, I only stumbled across the formula while blindly tinkering with reward functions for a productivity gamification system. In short, I needed a function $g(x)$ that behaves very similarly to $f(x)=x$ below a certain constant, e.g. 27 (which I can scale arbitrarily), then tapers off and flatlines. $g(x)=\log^e{(x+1)}$ happened to work perfectly, but of course I wouldn't assert the formula to be meaningful in any natural sense.

• The latter; the $e$th power. – Andrew Cheong Sep 29 '14 at 9:29
• @NajibIdrissi From the tries he put, I am convinced this is the $e$-th power of log, not the log in base $e$ – Martigan Sep 29 '14 at 9:29

## 1 Answer

Try the Lambert function, this is for such relations: http://en.wikipedia.org/wiki/Lambert_W_function

• Would you elaborate? I don't see a way to apply the $\mathrm{W}$ function here. – robjohn Sep 29 '14 at 12:43
• And if you take the exponentials of the terms? – Karl Sep 29 '14 at 17:42
• Try it; it doesn't work. Elaboration is more than asking another question. – robjohn Sep 29 '14 at 20:24