How can I know $\int_1^x\frac{dt}{t}$ is the inverse of exponential function?

Question

How can I know $\int_1^x\frac{dt}{t} \forall x>0$ is the inverse of exponential function assuming I've never heared of the natural logarithm.

Answer 1

Since $f(x)=e^x$ is a solution of the ODE $$ f' = f $$ and it is an increasing positive function, given that $g(x)$ is the inverse function of $f(x)$ we have: $$ g(f(x)) = x,$$ hence by differentiating we get: $$ f'(x)\, g'(f(x)) = 1, $$ or: $$ g'(f(x)) = \frac{1}{f'(x)} = \frac{1}{f(x)}, $$ $$ g'(t) = \frac{1}{t}. $$ Since $f(0)=1$ implies $g(1)=0$, the last line gives: $$ g(t) = \int_{1}^{t}\frac{du}{u}$$ as wanted. The domain of $g$ is the range of $f$, hence we must have $t>0$.

Answer 2

Let $f(x)=\int_1^x t^{-1}dt$, for $x>0$. Then, by FTC,

$$f'(x)=x^{-1}.$$

Then: $$\left(f({\rm e}^x)\right)'=f'({\rm e}^x){\rm e}^x={\rm e}^{-x}{\rm e}^x=1.$$

Hence: $f({\rm e}^x)=x+ C.$ As $f(1)=0$, we have $C=0$.

Answer 3

By the way: Looking at the involved areas under the curve $y={1\over t}$ one verifies easily that $$\int_1^{x\cdot y}{dt\over t}=\int_1^x{dt\over t}+\int_x^{x\cdot y}{dt\over t}=\int_1^x{dt\over t}+\int_1^y{dt\over t}\ .$$ This says that the function $$\ell(x):=\int_1^x{dt\over t}\qquad(x\geq1)$$ satisfies the functional equation of the logarithm; furthermore $\ell'(1)=1$.