Proof of natural log

Question

Suppose that $L:(0,\infty)\rightarrow \mathbb{R}$ is differentiable, that $L(1)=0$, and that $L'(x)=1/x$ for all $x>0$.

Then $L(2^n)>n/2$ for all $n\in N$.

This implies that $\sup(L)=\infty$, and a similar argument shows that $\inf L = -\infty$. By the inverse-function teorem, it follows that $L$ is a bijection from $(0,\infty)$ onto $\mathbb{R}$ and that its inverse $E=L^{-1}$ is differentiable. Show that $E'(t)=E(t)$ for all $t\in\mathbb{R}$.

Obviously the function is the natural log, but for this I can not use the properties of $\ln$ nor integration. So how do I show this without knowing what the function is or its inverse?

Answer 1

You don't necessarily have to find $E'(t)$ for this problem. Sometimes you can show two functions are equal without ever actually knowing what either function is.

Since $E$ is the inverse of $L$, for all $x \in (0, \infty)$ you have $E(L(x)) = x$.

Presuming you are allowed to use differentiation and the Chain Rule, you can take the derivatives of both sides with respect to $x$. You should then be able to relate $E'(L(x))$ to $E(L(x))$.

Edit: It occurred to me later that the solution might come even easier if we write $L(E(t)) = t$.

Answer 2

You know that $L$ is the logarithm, so that $L(2^n) = n L(2) > n/2$. Here is how it works: since $$ L(2^n) = \int_1^{2^n} L'(t) \, dt = \int_1^{2^n} \frac 1t \, dt$$ you can use the substitution $$s^n = t,\quad ns^{n-1} ds = dt, \quad \frac{dt}{t} = n\frac{ds}{s}$$ to get $$ \int_1^{2^n} \frac 1t \, dt = n \int_1^2 \frac 1s \, ds \ge n \int_1^2 \frac 12 \, ds= \frac n2.$$