I was following the book of Ross: Elementary analysis: the theory of calculus.

The exponential function is defined by power series: $e^x:=\sum_{n=0}^{\infty} x^n/n!$.

The logarithm function is defined using integration: for $x>0$, $\log x=\int_{1}^x \frac{1}{t} dt$.

Taking these as definitions, how can we show that these functions are inverses of each other in the sense $e^{\log x}=x$ and $\log e^x=x$?

A question with similar title has been posted earlier (see this), but there, the functions were possibly taken as inverses of each other, and then graphically it was interpreted what do it mean by inverses of each other.

Also in Lang's A first Course in Calculus defines logarithm as inverse of exponential.

However, I am taking above two definitions of these functions, and trying to show that they are inverses of each other. How to proceed? Any hint?


2 Answers 2


Hint: Use those definitions to show that both functions are differentiable, with $$ (e^x)' = e^x, (\log x)' = \frac 1x \, . $$ Then show that the function $f(x) =\frac 1x e^{\log x}$ (defined for $x> 0)$ has a zero derivative and is therefore constant: $f(x) = f(1) = 1$.

In order to show that $\log(e^x) = x$ you can proceed similarly, by considering $g(x) = \log(e^x) - x$.

If you know that $\exp$ and $\log$ are surjective (onto $\Bbb R_{>0}$ resp. $\Bbb R$) then it suffices to show one direction, and the other direction follows automatically.


Use these definitions to show that both are continuous, differentiable and strictly increasing on their domain and hence are invertible. Further the domain and range of these functions match like inverses.

Now you can note that if $y=e^x$ then $dy/dx=e^x=y $ or $dx/dy=1/y$ and then by integrating we get $x=\int_{1}^{y}\frac{dt}{t}=\log y$ and thus the desired conclusion is reached.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .