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:

$$b^x = e^{\ln (b) x}$$

But then somehow we get this really nice property:

$$b^x = \underbrace{b \cdot b \cdot \ldots \cdot b}_{x \text{ times}}$$ 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.

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I think exponentiation was first defined only for positive integers, then the definition was extended to integers, rationals and then reals, math.stackexchange.com/a/133238/164222. And so I think the approach given in your textbook is bad, since it consumes a lot of opportunities to let you understand things at an intuitive level. – Math Gems Aug 18 '14 at 23:50
You need to prove that $ln(x~y) = ln(x) + ln(y)$, which follows from the definition by a change of variable. Then it should not be hard as $e^x$ is defined by the inverse (in your text). – carlosayam Aug 18 '14 at 23:51
You can show rather easily through substitution that $\int_1^adt/t+\int_1^bdt/t=\int_1^{ab}dt/t$. The "nice property" can be derived by inverting this one (truly a slightly more complicated version of this one when the sum of two things is replaced by the sum of $n$ things). – PVAL Aug 18 '14 at 23:53

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 $\log x=\int_1^x\frac1t\,dt$ is a consequence of $(e^x)'=e^x$.

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.

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For the benefit of OP and others as well the "quite an insight to define exponentiation for all positive real numbers" is presented at paramanands.blogspot.com/2014/05/… Other approaches based on log as an integral are also discussed in detail. – Paramanand Singh Aug 19 '14 at 7:37
Nice! $\ \ \ \$ – Martin Argerami Aug 19 '14 at 13:45

You have $b^x=e^{x\ln{b}}$, and I presume you have $b^{x+y}=b^xb^y$, so:

\begin{aligned} b^x &=e^{x\ln{b}}\\ &=e^{\ln{b}+\ln{b}+\dotsb+\ln{b}}\\ &=e^{\ln{b}}e^{\ln{b}}\dotsm e^{\ln{b}}\\ &=b\cdot b \dotsm b \end{aligned}

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Replace most of your b with e and I agree with you! – Fezvez Aug 19 '14 at 0:15
I think you have a typo here. Your $= b^{xlnb}$ should read $= e^{xlnb}$. I would edit it, but I'm not sure about etiquette. – Nick R Aug 19 '14 at 0:15