# How can $1 + 2^{1/x} = 0$ be solved?

I'm trying to solve an equation:

$1 + 2^\frac{1}{x} = 0$

My first thought was to solve this by with the help of logarithms $\log_2 -1 = \frac{1}{x}$, but when I wrote it down, I saw a meaningless gibberish.

Then I tried to solve this, by representing -1 as a power of 2: $-\frac{2^1}{2^1}$

$1 + 2^\frac{1}{x} = 0 \\ 2^\frac{1}{x} = -\frac{2^1}{2^1} \\ 2^{\frac{1}{x} + 1} = -2^1$

But this is where I stuck and have no idea how to proceed. As far as I understand, powers are equal only if the bases are equal, but this is not the case here and there seems to be no way to get rid of this -1.

Wolfram Mathematica says, that there is no solutions in the Reals domain, but this doesn't eliminate the necessity to solve the equation.

Could anyone, please, at least suggest in which way should I proceed?

• There are obviously no real solutions, because $1 + 2^{1/x} > 1$ for all $x \ne 0$. There are complex solutions, because if you choose a branch of the logarithm, $x = \log(2) / \log(-1)$ is a solution.
– user296602
Nov 2, 2017 at 18:40
• There's no real solution since $2^{1/x}$ is always positive. For complex solutions you'll have to learn about the complex logarithm (en.wikipedia.org/wiki/Complex_logarithm) or wait for someone to write that kind of answer here. Nov 2, 2017 at 18:41
• If $a>0$, and $b$ is any real number then $a^b>0$. Therefore $1+a^b>1$. Nov 2, 2017 at 18:42

It is a famous identity that $e^{\pi i }= -1$. Now $2^{1/x}= e^{(\ln 2)/x}$.
So you want $\frac{\ln 2}{x} = \pi i$, and thus $x = -\frac{i\ln 2}{\pi}$ is a solution.
Other solutions are $-\frac{i\ln 2}{(2k+1)\pi}$ for integral $k$, and these are all (complex) solutions.
There are no real solutions, as $2^t$ for real $t$ is always positive.
since $1>0$ and $2^{1/x}>0$ we get no solutions