1. ## Property of Logs

This question has been bugging me lately

If a>0, use the functional equation for the logarithm to prove that log (a^r) = r log a for every rational number r.

2. Originally Posted by HallsFB
This question has been bugging me lately

If a>0, use the functional equation for the logarithm to prove that log (a^r) = r log a for every rational number r.
Hello,

I've attached a screenshot of my calculations:

3. Earboth only proved it for positive integers.
I will do it for rational numbers,

Let,
log (a^r) = n (1)
log a = m (2)

We want to show that,
n = r*m (3)

By definition equations (1) and (2) become,
10^n = a^r (4)
10^m = a (5)

Raise (5) to the power of r to get,
10^{r*m} = a^r
Now this is equal to (4) thus,
10^n = 10^{r*m}
Thus,
n = r*m
And we have shown (3)

4. Originally Posted by HallsFB
This question has been bugging me lately

If a>0, use the functional equation for the logarithm to prove that log (a^r) = r log a for every rational number r.
How about this one. It doesn't explicitly say r is rational as TPH's does, but I see no problem with it, since the proof should work for any real r.

Note: I will use the notation log[a]b to mean "log to the base a of b" where a and b represent any arbitrary variables or numbers.

We know by the definition of the logarithm that:

log[a]b = c
=> a^c = b

It seems that your problem is for log to the base 10, but it will work for any base, so i shall use dase x to represent any arbitrary base.

Proof:
We show log[x](a^r) = rlog[x]a

if r = 0, the proof is immediate and we have the desired result (since we would have log[x]1 = 0*log[x]a = 0). Now assume r is not zero

Let log[x](a^r) = y

=> x^y = a^r
=> (x^y)^(1/r) = (a^r)^(1/r) ........we can divide by r since it is not zero
=> x^(y/r) = a

rewriting this using the definition of logarithms again, we get:

log[x]a = y/r
=> rlog[x]a = y

But we said y = log[x](a^r)

Therefore, we have:
log[x](a^r) = rlog[x]a

5. Originally Posted by Jhevon
since the proof should work for any real r.
The definition for real exponents is:
Given a,b>0 we define:
a^b

As, exp(ln(a)*b)

The difficulty here is that we need to introduce the natural logarithm function ln x, and the its inverse function, the exponention e^x.

This is the difficutly with real exponents which is not covered in school.