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.
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)
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
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.