# How to solve $e^x = 2$

I know that $\ln(x)$ is the inverse of the exponential function $a^x$.

So I thought that

$$e^x=2 \Leftrightarrow x = \ln(2)$$

but my calculator says $x = \ln(2) + 2 i \pi n$, where $N \in \mathbb{Z}$. What have $e^x$ and $\ln(x)$ to do with the unit circle?

• The question is do you want to solve it for reals or for complex numbers? Commented Sep 18, 2013 at 14:36
• To see why $e^{2 \pi i n} = 1$ for all integers $n$, look up Euler's formula. Commented Sep 18, 2013 at 14:37
• I just have to ask what kind of calculator (not computer or website) gives that answer! Commented Sep 18, 2013 at 15:06
• @user40167, the long answer is "wikipedia", the short--"wiki". Commented Sep 18, 2013 at 16:49

This comes from the complex analysis ideas. If we know that $x$ is real valued, then clearly $x = \log 2$. However if $x$ is allowed to be complex valued, things become trickier. We know that for any $k\in\mathbb{Z}$, $\,\,e^{i\theta + 2k\pi i} = e^{i\theta}$. You can work this out yourself with Euler's formula. So the issue is that any complex number has infinitely many equivalent expressions (when you write it in polar form). What this boils down to is the fact that if you write your complex number in polar form, if you add $2\pi$ to the angle, you end up at the same exact complex number. Since we don't know exactly which range of $\theta$ you wanted for the complex number, we can't specify one value of $\theta$. So take your example.

$$e^x = e^{x+2k\pi i} = 2.$$

Let's rewrite $2$ slightly as $e^{\log 2 + 2l\pi i}$ for $l\in\mathbb{Z}$ and move it over to the left. We get

$$e^{x+2k\pi i - \log 2 - 2l\pi i} = 1 = e^{2m\pi i}$$

And so we know that $x+2k\pi i - \log 2 - 2l\pi i = 2m\pi i$ for some integer $m$. Since we don't care about the individual values of $k,l,m$ let's forget about them and just group them together. What we end up with is $x = \log 2 + 2n\pi i$ like you have.

Ostensibly the problem is that since there is no unique representation for a complex number in polar form (which implicitly invokes the complex exponential), there cannot possibly be any unique way to take logarithms. Inverses of multivalued functions are ill-conceived. To get around this, what we usually do is a priori restrict ourselves to certain ranges of $\theta$ (say $[0,2\pi)$). In this case, there is a unique polar expression and we can easily do the inversion.

• Fantastic answer! Commented Oct 1, 2015 at 11:02

In Complex Analysis one defines for $z=a+bi\in\mathbb{C}$ $$e^{z}:=e^{a}(\cos(b)+i\sin(b))$$

Then you can prove that for any $z_{1},z_{2}\in\mathbb{C}$ $$e^{z_{1}}\cdot e^{z_{2}}=e^{z_{1}+z_{2}}$$

Note that by this definition $$e^{2\pi in}=1$$

for all $n\in\mathbb{N}$.

Hence $$e^{\log(2)+2\pi in}=e^{\log(2)}\cdot e^{2\pi in}=2\cdot1=2$$

Euler's formula says that $$e^{i \theta}=\cos(\theta)+i\sin(\theta) \quad(*)$$ so let $\theta=2\pi n$ then $(*)$ becomes $e^{i2\pi n}=1$