glutynowy44

2022-07-16

If $\mathrm{ln}x$ is defined via an integral and $e$ defined from $\mathrm{ln}x$, how would you prove that $\mathrm{ln}x$ is the inverse of ${e}^{x}$?
This is a somewhat technically specific question about the relationship between $\mathrm{ln}x$ and ${e}^{x}$ given one possible definition of $\mathrm{ln}x$
Suppose that you define $\mathrm{ln}x$ as
$\mathrm{ln}x={\int }_{1}^{x}\frac{dt}{t}$
$\mathrm{ln}x={\int }_{1}^{x}\frac{dt}{t}$
We can use the connection between the integral definition of $\mathrm{ln}x$ and the harmonic series to show that $\mathrm{ln}x$ grows unboundedly. This function is monotonically increasing, so it should have an inverse mapping from its codomain ($\mathbb{R}$) to its domain (${\mathbb{R}}^{\mathbb{+}}$). Let's call that function ${\mathrm{ln}}^{-1}x$
Since $\mathrm{ln}x$ grows unboundedly and is zero when $x$ is $1$, we can define $e$ as the unique value such that $\mathrm{ln}e=1$
So here's my question: given these starting assumptions, how would you prove that ${\mathrm{ln}}^{-1}x={e}^{x}$ (or, equivalently, that $\mathrm{ln}x={\mathrm{log}}_{e}x$? I'm having a lot of trouble even seeing how you'd get started proving these facts, since basically every calculus fact I know about $e$ and $\mathrm{ln}x$ presumes this result to be true.
Thanks!

1s1zubv

Expert

If $f\left(x\right)=\mathrm{log}x$ is defined as a primitive of $\frac{1}{x}$ for which $f\left(1\right)=0$, then $f\left(ab\right)=f\left(a\right)+f\left(b\right)$ holds for any couple of positive real numbers. This gives that the inverse function $g\left(x\right)$ is a ${C}^{1}$ function, satisfies $g\left(0\right)=1$ and the functional equation:
$\begin{array}{}\text{(1)}& g\left(a\right)g\left(b\right)=g\left(a+b\right)\end{array}$
for any couple of real numbers $a,b$. $\left(1\right)$ and differentiability grants that ${g}^{\prime }=g$, hence $g\left(x\right)={e}^{x}$

Banguizb

Expert

The advantage of defining $\mathrm{ln}x$ as an integral is that it gives a simple way to define ${e}^{x}$ in elementary calculus (without resorting to infinite series, taking limits of rational powers of $e$, or other methods).
Once $\mathrm{ln}x$ has been defined in this way, you can establish that $y=\mathrm{ln}x$ is an increasing function whose range is $\mathbb{R}$, so it has an inverse function with domain $\mathbb{R}$ which we can $\mathbf{d}\mathbf{e}\mathbf{f}\mathbf{i}\mathbf{n}\mathbf{e}$ to be $y={e}^{x}$
With this definition, we have the identities
$\mathrm{ln}\left({e}^{x}\right)=x$ for all $x$ and ${e}^{\mathrm{ln}x}=x$ for $x>0$, and $\mathrm{ln}e=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{e}^{1}=e$

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