Question

Prove that, for all non-zero a and b,\[(ab)^x = a^x * b^x\] Prove that, for all non-zero a and b,\[(ab)^x = a^x * b^x\] @Mathematics

Answer 1

Depends on your definition of \( x^y \).

Answer 2

true, which is why I need help with that :(

Answer 3

It's a bit messy and truly depends on where you are, so I'm quite serious as to what is your defintion of x^y

Answer 4

I don't know the best definition... if I define it as a product using that big pi notation, I can't get all reals. If I use the exp() and ln() functions I only get the positive reals. I want all the reals to work.

Answer 5

I'm sure there's a straightforward, albeit tedious, way to fit all the non-zero reals into the a and b, and all reals into the x

Answer 6

isn't if sufficient proof to just state:\[(ab)^x=(ab)*(ab)*(ab)*...\]so you get 'x' lots of multiplications and then just use the commutative and associative properties to re-arrange this as:\[=(a*a*a*...)*(b*b*b*...)\]and hence:\[=a^x*b^x\]

Answer 7

if == it

Answer 8

what is '1/4' lots of multiplications, or '-1' lots if we do it that way? :(

Answer 9

hmmm... I see professor that you are playing with my mind :-)

Answer 10

I will keep this question to annoy the math faculty.

Answer 11

along with the torus cutting problem and some other ones

Answer 12

In the most general form, we define exponent in the complex numbers and then restrict back to the real:

Let z be a complex number and Arg(z) be defined as its principle argument, \(Arg(z) \in [0,2\pi) \)

Then if ln is the usual real valued function for ln, we define the Ln(z) to be
\[ Ln(z) = ln |z| + Arg(z) \]
and another 'function', although it is actually not a function,
\[ ln(z) = ln |z| + Arg(z) + 2k\pi, \ \ k \in Z \]

Now, with that definition, we can make sense of \( w^z \) for any complex number \( z \) and \( w \neq 0 \):

\[ w^z = exp(z . ln(w)) \]
Notice by construction w^z has nearly always infinitely many values. We say the principle value is
\[ exp(z . Ln(w)) \]

Suppose now z and w are real numbers and w is not zero. Then w^z is the principle value of w^z considered as complex numbers provided that number is real. If it is not, then w^z is not defined.

Finally, there is the issue of w = 0. Usually (but not always!) it is easiest to say that 0^z is zero provided Re(z) > 0 and undefined otherwise.

Answer 13

Now with that definition, you can show your result if you're careful.

Answer 14

I'll print that answer so once that Spivak book teaches me what a complex number is, my mind will be able to tackle this :-D

Answer 15

aha, could we use logs, let:\[n=(ab)^x\]\[\ln n=\ln (ab)^x=x\ln(ab)=x(\ln(a)+\ln(b))=x\ln(a)+x\ln(b)=\ln a^x+\ln b^x\]\[=\ln (a^x*b^x)\]therefore:\[n=a^x*b^x\]so:\[(ab)^x=a^x*b^x\]

Answer 16

if we use logs, then we have problems when either a or b (and perhaps both at once) are negative

Answer 17

When you do you'll see that exponents can become multi-valued very often and it leads to a radical change in the way we think of the domain of complex-valued functions, a topic called "Riemann surfaces".

Answer 18

I will ask my calculus II professor what a Riemann surface is :-D

Answer 19

hmmm - this maths beast is a lot more subtle than I had envisaged :-(
I guess you guys are working at a much higher plateau than I am at!

Answer 20

lol. He's either going to love you or hate you.

Answer 21

Alright; I can sleep now that I know that we can change (ab)^x into a^x * b^x