# Other log solutions?

I am evaluating the expression:

$\ln(1)$

And I know the trivial solution is $0$.

Are there other solutions to this equation? I feel there should be, my logic is as follows:

if:

$\ln(1) = x \implies 1 = e^x \implies 1 = 1 + x + x^2/2! + x^3/3!...$ $\qquad\qquad\qquad\qquad\qquad\implies 0 = x + x^2/2! + x^3/3!...$

$\implies x = 0$ is one solution, the other solution is all $x$ such that: $$1 + x/2! + x^2/3! + x^3/4! ... = 0$$

There have to be other solutions, or limiting solutions...

What are they?

• Limiting solutions are basically directions which when continued out to infinity tend to zero. Example: 1/x has the entire unit circle as a limiting solution: where by picking a point on the unit circle (in the complex plane) and continuing out towards infinity you will tend to zero. Commented May 27, 2013 at 3:49

There is no equation mentioned in the post. I assume you mean the equation $e^x=1$. This has no real solutions other than $x=0$. You can see this from the fact that the function $e^x$ is an increasing function, so can take on the value $1$ at most once.
The logarithm function can be extended to the complex numbers, where it is multiple-valued. In the complex numbers, the equation $e^z=1$ has the solutions $z=2n\pi i$, where $n$ ranges over the integers. So in the complex numbers, $\ln 1$ takes on infinitely many values.
• The first paragraph dealt with real numbers, so I called the variable there $x$. The second paragraph dealt with the complex case, so there I used the more traditional variable name $z$. Of course $x$ or $y$ would be not unsuitable, though being traditionally minded, if I did not use $z$ my second choice would be $w$. Commented May 27, 2013 at 16:43
If $x$ is real, $1=e^x$ has the single solution $x=0$. The easiest way to see this is that $e^x$ is monotonic, so there can only be one solution. In the complex plane, because $e^{2\pi}=1$, we can solve $1=e^x$ with $x=2k\pi i$ for $k$ integral.