---
tag: A49EDF2F52
kind: proof
name: Proof that the logarithm is the inverse of the exponential
msc: [33B10]
proves: 0RKW2X63YR
uses: [0RMVS7RXZX, 0044CPHJBV, 0EX9GFJ0B3, 07ZMX4ANJS, 0P9BNBW2W3, 0KN2V7840E, 0E8HNVBCSP, 012C7GFF20, 00KAFBR32E, 07BY47HG5S]
sources:
  - url: https://en.wikipedia.org/wiki/Natural_logarithm
    locator: Definition and properties
---

## Proof

We show that $\exp$ is a continuous strictly increasing bijection from $\R$ onto $(0, \infty)$, and that $\ln$ is its inverse.

1. Write $e^y = \exp(y)$ for the [[exponential function|0RMVS7RXZX]]. It is differentiable by [[its derivative rule|0EX9GFJ0B3]], so it is [[continuous|07ZMX4ANJS]].
2. It is [[positive|0P9BNBW2W3]] and [[strictly increasing|0KN2V7840E]]. So it maps $\R$ into $(0, \infty)$, and it is injective.
3. By the [[exponential of a sum|0E8HNVBCSP]], $e^y e^{-y} = e^0 = 1$, so $e^{-y} = 1/e^y$.
4. Let $x > 0$. We find $y$ with $e^y = x$. By the inequality [[$1 + t \le e^t$|012C7GFF20]], $e^x \ge 1 + x > x$.
5. By steps 3 and 4, $e^{-1/x} = \frac{1}{e^{1/x}} \le \frac{1}{1 + 1/x} = \frac{x}{x + 1} < x$.
6. By steps 4 and 5, $e^{-1/x} < x < e^x$. Also $-1/x < 0 < x$.
7. By the [[intermediate value theorem|00KAFBR32E]] on $[-1/x, x]$, there is $y$ with $e^y = x$. By step 2, this $y$ is unique.
8. So $\exp$ is a bijection from $\R$ onto $(0, \infty)$. For $x > 0$, the [[natural logarithm|0044CPHJBV]] $\ln x$ is the unique $y$ of step 7. So $\ln$ is the inverse function of $\exp$.
9. For $x > 0$, $e^{\ln x} = x$ by the definition of $\ln x$.
10. For real $y$, the number $e^y$ is positive, and $y$ is a solution of $e^t = e^y$. By uniqueness in step 7, $\ln(e^y) = y$.
11. By the theorem on the [[continuity of the inverse|07BY47HG5S]] with $I = \R$ and $f = \exp$, the inverse $\ln$ is continuous and strictly increasing on $\exp(\R) = (0, \infty)$.
12. By step 10, each real $y$ equals $\ln(e^y)$, so the range of $\ln$ is $\R$.

## Idea

The exponential function is continuous and strictly increasing. It takes values larger than any given positive $x$ and smaller than $x$, so it hits $x$ exactly once. Its inverse is $\ln$, and the inverse of a continuous strictly increasing function on an interval is again continuous and strictly increasing.