Proof of exponentiation property

[mnb96]

Hello,
this might be a silly question for many of you.
How would you prove that:

$$(a^{x})^y = a^{x\cdot y}$$

when a,x,y are reals and a>0. The case for x,y integers is easy to prove, but how would you extend the proof to real numbers?

 

[tiny-tim]

hello mnb96!

if a = e, then it's obvious from the definition (e^x = ∑ xⁿ/n!) …

just write out the two expansions, and multiply them, term-by-term

if a ≠ e, use a^x = e^xln(a)

however, if you object to this method, then how are you defining a^x with x irrational?

if you're using a limit definition (of a sequence of rational numbers converging to x), then use the same sequence to prove the formula
​

 

[mnb96]

Thanks tiny-tim!
I was just thinking that you used the Taylor expansion of e^x, and as a consequence, also the concept of derivative of e^x (and finally also the logarithm).

Is all that stuff required? or is it possible to avoid it?

Moreover, when you write a^x = e^xln(a) aren't you implicitly assuming that e^xln(a)=(e^x)^ln(a) which is what we wanted to prove?
EDIT: ops sorry...we could indeed assume in that case that we already proved the statement for a=e.

 

[tiny-tim]

Thanks tiny-tim!
I was just thinking that you used the Taylor expansion of e^x, and as a consequence, also the concept of derivative of e^x.

Is all that stuff required? or is it possible to avoid it?

you don't need the derivative at all, just multiply the two expansions

 

[mnb96]

you don't need the derivative at all, just multiply the two expansions

Ah, but so how did you manage to write e^x = ∑ xⁿ/n!
It popped out from nothing without using the Taylor expansion?

EDIT: oh! I found it http://en.wikipedia.org/wiki/Charac...ion#Equivalence_of_characterizations_1_and_2"...not so trivial after all, if we just start from the definition of the number e

 

[lurflurf]

(1) (a^x)^y=a^{x y}

How one would prove (1) would depend upon the definition one would take. Really one would like (1) to be true and choose a definition to make it true.

I like the definition u^v:=e^{v log(u)}

 

[robert Ihnot]

When I took the Calculus, it was the difinition of e^x, that is as x goes to infinity.

Ah, but so how did you manage to write e^x = ∑ x^n/n! It popped out from nothing without using the Taylor expansion?

 

[tiny-tim]

Ah, but so how did you manage to write e^x = ∑ xⁿ/n!
It popped out from nothing without using the Taylor expansion?

that is the taylor expansion of e^x, see http://en.wikipedia.org/wiki/Taylor_expansion#Examples"

 

[HallsofIvy]

I keep pointing out to people that how you prove something like this depends heavily upon how you define the terms. And there may be sevaral different ways of defining the same thing.

It is, as you said, easy to show that, for n and m positive integers, $a^xa^y= a^{x+y}$ and $(a^x)^y= a^{xy}$, for a any positive number, in particular, e. One method of proceeding is to then define $a^x$, for x not a positive integer, so that those formulas are true. For example, if y= 0, $a^xa^y= a^{x+ y}$ becomes $a^xa^0= a^{x+ 0}= a^x$. From that, for a any non-zero number, it follows that defining $a^0= 1$, we are "preserving" that identity. Similarly, any negative integer can be written as -n for n a positive integer. With x= n, y= -n, we have $a^na^{-n}= a^{n-n}= a^0= 1$ so, again, for a non-zero, we must have $a^{-n}= 1/a^n$ and so we can define a to a negative integer power, $a^{-n}$, to be the reciprocal of [/itex]a^n[/itex] in order to keep that useful formula true.

If n is a non-zero integer, then $(a^{1/n})^{n}= a^{n(1/n)}= a^1= a$. That is, in order that that "law of exponents" be true even for non-integer powers, we must have $a^{1/n}$ equal to a number whose nth power is a. In order to be sure such a thing exists (I am assuming real numbers here) we must require that a be positive. In that case, there may be two such real numbers. We define $a^{1/n}$ to be the positive such root. Of course, it follows that $a^{m/n}= (a^{1/n})^m$ so we use that to define a^r for any rational number.

Finally, we define a to an irrational power by requiring that $a^x$ (again, for a any positive number and so, in particular, e^x) be continuous. That is, if $\{r_1, r_2, r_3, \cdot\cdot\cdot, \}$ is a sequence of rational numbers converging to x, then we define $a^x$ to be the limit $\displaytype\lim_{n\to\infty} a^{r_n}$.

That way we can assert that both $a^xa^y= a^{x+y}$ and $(a^x)^y= a^{xy}$ are true by definition.



For a completely different point of view, and a completely different proof, we can define $$e^x$$
to be the inverse function to ln(x) while defining ln(x) itself by
$$ln(x)= \int_1^x \frac{1}{t}dt$$

Let x be a postive real number, y any real number. Then, by that definition,
$$ln(x^y)= \int_1^{x^y} \frac{1}{t}dt$$

If $y\ne 0$, let $u= t^{1/y}$. Then $t= u^y$ so that [itex]dt= y u^{y-1}dy[/tex]. When t= 1, u= 1 and when $t= x^y$, u= x. So with that change of variable, the integral becomes
$$ln(x^y)= \int_1^x \frac{1}{u^y}(yu^{y-1}dy)= y\int_1^x \frac{1}{u} du= yln(x)$$.

If y= 0, then $x^y= x^0= 1$ so that
$ln(x^y)= ln(1)= \int_1^1\frac{1}{t}dt= 0$
so we still have $ln(x^y)= ln(1)= 0= 0(ln(1))= yln(x)$.

Now, if we let $w= (e^x)^y$, with $y\ne 0$, we have $w^{1/y}= e^x$ and, since $e^x$ is defined as the inverse function to ln(x), we have $x= ln(w^{1/y})= (1/y)ln(w)$ and so $xy= ln(w)$. Now return to the exponential form:
$w= e^{xy}$ proving that $w= (e^x)^y= e^{xy}$.

Of course, if y= 0, then $(e^x)^y= (e^x)^0= 1= e^0= e^{x(0)}= e^{xy}$