Isolate x in an exponential function?

Question

I have $x=2^x-1$. How do I isolate x? I tried to take the log of both sides of $x-1=2^x$, but then there's a $ln(x-1)$.

Answer 1

There are two trivial solutions: $x=0,x=1$. There are no other real solutions.

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For complex numbers:

Solving with the Lambert W function, I get

$$x-1=2^x$$

$x-1=u$

$$u=2^{u+1}$$

$$u=2e^{\ln(2)u}$$

$$ue^{-\ln(2)u}=2$$

$$-\ln(2)ue^{-\ln(2)u}=-2\ln(2)=-\ln(4)$$

$$-\ln(2)u=W(-\ln(4))$$

$$u=\frac{W(-\ln(4))}{-\ln(2)}$$

$$x=\frac{W(-\ln(4))}{-\ln(2)}+1$$

where $W(z)$ is the inverse of $ze^z$ and $e$ is Euler's number. Choosing different branches will give you different answers, and two of the branches should admit $x=0,x=1$.

Answer 2

Already said, but if you are supposed to find the solutions to the equation: Show that $2^x-(1+x)$ is strictly convex (2nd derivative strictly positive), that there are two evident solutions and conclude.