I was just thinking, how is y^x defined when x is irrational (i.e. you can't say it's the q-th root of y to the power of p, if x = p/q). Also, why is it that x^1/q = the q-th root of x? It's probably defined that way, but you could define it however you wanted - what's the reasoning behind it? If that makes sense.

Regards,
Olof

Hi Olof, there's various ways of doing it (as always).

You can define y^x to be e^{x log(y)} which deals with rationals, irrationals, complex numbers, or whatever you like. If you define it that way you have to show that this corresponds to your definitions of y^x when x is rational, but this isn't too difficult.

The other way of doing it is to define it as a limit. So, if the sequence of rationals p_n /q_n tends to x as n tends to infinity (and every irrational has such a sequence) then we define y^x to be the limit of y^pn/qn as n tends to infinity. The reason for defining x^1/q to be the q-th root of x is because we want to preserve the idea that (x^a )^b =x^ab for all a and b. If we want this relationship to hold for all a and b, then it a=1/p and b=p we want (x^1/p )^p =x, which just says x^1/p is the p-th root of x. Does that help?

I see Dan, thanks a lot. I suppose that the relationship (x^a )^b =x^ab comes from logs? How would one show that log ab = log a + log b? Am I right in thinking this can be used to show that log aⁿ = n log a, which would then show (x^a )^b =x^ab ?

Thanks again,
Olof

Well, (x^a )^b =x^ab can be proved for integer a,b without using logs, and this could be a reason for thinking that it should also be true if a and b are nonintegers.

Alternatively you can do it usings logs. You can prove that log ab = log a + log b by saying that ab=e^log(ab) , a=e^log(a) , b=e^log(b) , so e^log(ab) =e^log(a) e^log(b) =e^{log(a)+log(b)} , taking logs we get log(ab)=log(a)+log(b).

There are various starting points for all of these things. If you've defined x^y to be e^ylog(x) then obviously log(x^y )=log(e^ylog(x) )=ylog(x). However, if you've defined x^y in another way you have to show that x^y =e^ylog(x) , and so on. We can prove (x^a )^b =x^ab for integer a and b just by noting that (x^a )^b =x^a x^a ...x^a (b times), and so =x^a+a+...+a =x^ab .

There is another way of proving that $\ln ab=\ln a+\ln b$, which should show that $\log ab=\log a+\log b$, though I haven't given this much thought.

It uses integration of the curve of $y=1/x$.

If we integrated between the limits of 1 and $a$, where $a$ is greater than 1.

${\int }_1^a(1/x)\mathrm{dx}=\ln a=A_1$

Call this $A_1$ . $b$ is another $x$-value greater than 1. We integrate between 1 and $ab$

${\int }_1^{ab}(1/x)\mathrm{dx}=\ln ab=A_3$

We call this $A_3$ since the area between $a$ and $ab$ is $A_2$. It can be seen that $A_1+A_2=A_3$

We cannot simply integrate $y$ between $a$ and $ab$ as our answer won't be in terms of $b$ only.

If we use the substitution $x=au$, where $u$ is a function of $x$ then $\mathrm{dx}=a\mathrm{du}$

The only way that $x=au$ is if

$a=1$ and $ab=b$

To find $A_2$

${\int }_a^{ab}(1/x)\mathrm{dx}$

Using the substitutions $a=1$ $ab=b$ $x=au$ $\mathrm{dx}=a\mathrm{du}$

$A_2={\int }_1^b(1/au)a\mathrm{du}={\int }_1^b(1/u)\mathrm{du}=\ln b$

Since $A_1+A_2=A_3$

then $\ln a+\ln b=\ln ab$

Thanks Ali and Dan. I hadn't thought of it that way before Ali - thanks for pointing it out.

One last thing - is it as easy to show that x^a x^b = x^a+b , for real a & b? I'll have a go in a moment.

/Olof

Just use the definition Dan gave.

Kerwin

Do you do more rigorous proofs of this kind of thing at university?

Olof, yes, when proving it rigorously you have to worry about convergence, that's the main problem (which is true if you define y^x as the limit of y^pn/qn , or if you define y^x as e^xlog(y) , since you have to prove that exp and log are well defined functions, lots of work on power series and stuff).