How to prove that $(x^a)^b=x^{ab}$

Question

Let $a$ and $b$ be rational numbers. How can I prove that $(x^a)^b=x^{ab}$. I just don't know where to begin, can someone tell me a hint? I know that if $a=c/d$ and $b=e/f$ then $(x^a)^b$ is ${(x^{c/d})}^{e/f}$ but then what can I do?.

Answer 1

You would need to start by proving that $(x^n)^m = x^{nm}$ whenever $n, m$ are natural numbers. To do this, proceed by induction on $m$.

You would then need to extend this to the theorem that $(x^n)^m = x^{nm}$ whenever $n, m$ are integers (and both sides of the equality are defined). To do this, you need to do case analysis on the signs of $n$ and $m$ and cite the above theorem.

Then, you should extend this to rational numbers. Recall that given a rational number $\frac{n}{m}$ and a positive number $x$, the quantity $x^{n/m}$ is the unique positive number satisfying $(x^{n/m})^m = x^n$. Using this definition, you should find it straightforward to extend the previous result to the rationals.

Answer 2

The proof depends on your definition of exponentiation. If you define $x^y$ as $\textrm{exp}(y \ln x)$ then $$ (x^a)^b=\exp({b\ln(\exp (a\ln x))})=\exp(ba \ln x)=x^{ab}. $$ If instead you define $x^{c/d}$ as $\sqrt[d]{x^c},$ then using the version of the law for integers, $$ ((x^{c/d})^{e/f})^{df}=\left(\left(\sqrt[f]{\left(\sqrt[d]{x^{c}}\right)^e}\right)^{f}\right)^d=((\sqrt[d]{x^c})^e)^d=((\sqrt[d]{x^c})^d)^e=x^{ce}, $$ therefore $$ (x^{c/d})^{e/f}=\sqrt[df]{x^{cd}}=x^{\frac{c}{d}\cdot\frac{e}{f}}. $$

Answer 3

If $x\in\mathbb R^{+}$ and ${\frac ab}$ is a rational number, with $a,b\in\mathbb Z$, then $x^{\frac {a}{b}}$ is defined as:

$${x^{\frac {a}{b}}=\left(x^{a}\right)^{\frac {1}{b}}=\left(x^{\frac {1}{b}}\right)^{a}}$$

The equality on the right may be derived by putting $y=x^{\frac {1}{b}}$ and writing

$$\begin{align}\left(x^{\frac {1}{ b}}\right)^{a}&=y^{a}\\ &=\left((y^{a})^{b}\right)^{\frac {1}{b}}\\ &=\left((y^{b})^{a}\right)^{\frac {1}{b}}\\ &=\left(x^{a}\right)^{\frac {1}{b}}.\end{align}$$

If $r\in\mathbb Q^{+}$, then $0^r=0$ by definition.

All these definitions are required for extending the identity $$\left(x^a\right)^{b}=x^{ab}$$ to rational exponents.