detineerlf

2022-07-21

How does exponentiation relate to multiplication?
My book derives the logarithm function as a definite integral of $1/x$ and defines the exponential function as its inverse. It then extends this definition to other bases:
${b}^{x}={e}^{\mathrm{ln}\left(b\right)x}$
But then somehow we get this really nice property:

if $x$ is an integer.
This property is too nice to just be a coincidence. Is there a deeper reason this property emerges? I mean it's hard to see how functions derived from relatively complex operations on $1/x$ (integration) would somehow simplify to such a simple definition for integer arguments.

phinny5608tt

Expert

Of course it is not a coincidence. Historically, you first define ${b}^{n}$ for $n$ a positive integer, and you go from there. Thing is that with this approach you need quite an insight to define exponentiation for all positive real numbers, and to deduce the existence and properties of the exponential function. You then define the logarithm as its inverse, and the equality $\mathrm{log}x={\int }_{1}^{x}\frac{1}{t}\phantom{\rule{thinmathspace}{0ex}}dt$ is a consequence of $\left({e}^{x}{\right)}^{\prime }={e}^{x}$
Starting the opposite way, by defining the logarithm, is not as intuitive but leads to much easier proofs. In particular there is no struggle to define the exponential, as it is simply the inverse function of the log.

Damien Horton

Expert

You have ${b}^{x}={e}^{x\mathrm{ln}b}$, and I presume you have ${b}^{x+y}={b}^{x}{b}^{y}$, so:
$\begin{array}{rl}{b}^{x}& ={e}^{x\mathrm{ln}b}\\ & ={e}^{\mathrm{ln}b+\mathrm{ln}b+\cdots +\mathrm{ln}b}\\ & ={e}^{\mathrm{ln}b}{e}^{\mathrm{ln}b}\cdots {e}^{\mathrm{ln}b}\\ & =b\cdot b\cdots b\end{array}$

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