glutynowy44

Answered

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!

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!

Answer & Explanation

1s1zubv

Expert

2022-07-17Added 17 answers

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

$\begin{array}{}\text{(1)}& g(a)g(b)=g(a+b)\end{array}$

for any couple of real numbers $a,b$. $(1)$ and differentiability grants that ${g}^{\prime}=g$, hence $g(x)={e}^{x}$

$\begin{array}{}\text{(1)}& g(a)g(b)=g(a+b)\end{array}$

for any couple of real numbers $a,b$. $(1)$ and differentiability grants that ${g}^{\prime}=g$, hence $g(x)={e}^{x}$

Banguizb

Expert

2022-07-18Added 3 answers

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

$\mathrm{ln}({e}^{x})=x$ for all $x$ and ${e}^{\mathrm{ln}x}=x$ for $x>0$, and $\mathrm{ln}e=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{e}^{1}=e$

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

$\mathrm{ln}({e}^{x})=x$ for all $x$ and ${e}^{\mathrm{ln}x}=x$ for $x>0$, and $\mathrm{ln}e=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{e}^{1}=e$

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