# Prove by Induction that $e^{nx}=(e^x)^n$

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:

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

Part B:

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

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Do you know what the structure of a proof by induction is? Did you check the base case, and try to set up the inductive step? I suggest using $(n+1)x=nx + x$ and $0 = nx - nx$ for parts A and B, respectively. –  Jonas Meyer Dec 14 '12 at 6:36
Hint: Part A: $E((n-1)x+x) = E((n-1)x)E(x)$ Part B: Prove E(0) = 1 and then from part A, what is $E(nx+ (-nx))$? –  Isomorphism Dec 14 '12 at 6:37
Please do not delete you question. This posting could be helpful to someone else in future –  user127.0.0.1 Dec 17 '12 at 23:58

The statement is clearly true for n=1. Assume it to be true for n=k, i.e., $E(kx)=E(x)^k$. Then the statement follows for n=k+1 as $E((k+1)x)=E(kx+x)=E(kx)E(x)=E(x)^k.E(x)$, and we are done.
Part B follows immediately because $E(0)=E(-nx+nx)=E(-nx)E(nx)$. But $E(0)=1$ as $L(1)=0$.
Alternatively, $E(0)=1$ follows from $E(x+0)=E(x)E(0)$. –  Mario Carneiro Dec 14 '12 at 8:35