# Taking $e^{xn}=(e^x)^n$ a step further to rational numbers.

Working on an assignment but I've run into a stumbling block! I've got a couple of problems that I don't know how to do! The problems are attempting to have you define the $\log$ and $\exp$ functions. Thanks in advance! I can't wait to be done with this Analysis course!

For any $x \in (0,\infty)$, define $L(x)=\int_{1}^x {1\over t} dt$

$L(1/x)=-L(x)$

$L(x)$ is invertible and its inverse is $E(x)$

$E'(x)=E(x)$

$L(ax)=L(x)+L(a)$

$E(y+z)=E(y)E(x)$

Part A: Done

Let $n$ be a positive integer. Prove by induction that $E(nx)=E(x)^n$.

Part B: Done

Deduce from (a) that we also have $E(-nx)=E(x)^{-n}$, so (a) holds for all integers $n$.

Part C: My Problem Child

Deduce (give details) that for any rational number $r={m\over n},$with $n, m$ integers, $m>0$, that $E(rx)=E(x)^r$

For this last part, I feel like I can just say something like let $r=n$, then the truth of Part A applies Part C, which implies that: $E(rx)=E(x)^r$, but that seems way to easy!

Thanks for the help!

-
It's just that easy (specifically, $E(nx)=E(x)^n\Rightarrow E(y)=E(y/n)^n$). The fun part is the next step, where you generalize from rational numbers to reals! –  Mario Carneiro Dec 16 '12 at 23:22
$$E\left(n\left(\frac{mx}{n}\right)\right)=E\left(\frac{mx}{n}\right)^n$$ $$E\left(\frac{mx}{n}\right)=E\left(n\frac{mx}{n}\right)^{1/n}=E(mx)^{1/n}=\left(E(x)^m\right)^{1/n}=E\left(x\right)^{m/n}$$
I think you want multiplication in the $E(\,\cdot\,)$... –  Mario Carneiro Dec 16 '12 at 23:24