Can the Power Rule be applied to all rational numbers in logarithms?

[brh2113]

All information, including the problem, is attached. So far I think I've proven by induction that log (a^r) = r log (a) whenever r is an integer, but I need to prove this for all rational numbers r = p/q.

We're working with the functional equation that has the property that f(xy) = f(x) + f(y), and we're supposed to prove the equality using this. My initial thoughts were to write f(x*x^{p/q - 1}) = f(x) + f(x^{p/q - 1}), but it didn't get me anywhere. Any thoughts or suggestions?

 

[rock.freak667]

Let log_z b^n =z
a^z=b^n
a^{\frac{z}{n}}=b
and by the definition of logs
\frac{z}{n}=log_a b
then multiply by n

 

[HallsofIvy]

You should have said in your post what you say in your attachment: that you are to use the "functional equation" log(xy)= log(x)+ log(y) to prove that log(a^r)= r log(a).
Yes, you can prove that log(aⁿ)= n log(a) for any positive integer by induction. Also if n= 0, then a⁰= 1 so log(a⁰)= 0= 0log(a).

If m is a negative integer, then there exist a positive integer n such that m+n= 0.
log(a^m+n)= log(aⁿ)+ log(a^m). With m+n= 0, what does that tell you.

Now, go back and prove that log(a^nx)= n log(a^x) in exactly the same way (or include x from the start) for x any real number. What happens if x= 1/n?

 

[brh2113]

Ah I see. Since

log (a^{m+n}) = log (a^{0}) = 0 = log (a^{n}) + log (a^{m}),

log (a^{n}) = log (a^{m}). This implies that nlog(a) = (-m)log(a), which means that the formula is true for all positive and negative integers, plus zero. Right?

Now with r = p/q, I can write f(x^{p/q}) = f(x^{p*(-q)}), at which point I can say that because both p and q are integers I have

f(x^{p*(-q)}) = (-p/q)f(x).

Oh wait. I've just realized that f(x^{-n}) \neq (-n)f(x). The negative sign shouldn't be there: I should be trying to get

f(x^{-n}) = (1/n)f(x).

Now I'm completely lost. I think I went wrong with the true statement
nlog(a) = (-m)log(a), because I forgot that (-m) is positive. After this error I can't seem to get back on track. Help?

EDIT: Looking at it again, I've realized another mistake:
f(x^{p*(-q)}) = pf(x) - qf(x), not what I stated before.

 

[HallsofIvy]

Ah I see. Since

log (a^{m+n}) = log (a^{0}) = 0 = log (a^{n}) + log (a^{m}),

log (a^{n}) = log (a^{m}).

Uhhh, no. Assuming, from "a^m+n= a⁰", you mean m= -n, this is correct until the last line which should be log(aⁿ)= -log(a^m). That may be just a typo since you have the negative in the next line.

This implies that nlog(a) = (-m)log(a), which means that the formula is true for all positive and negative integers, plus zero. Right?

Yes.

Now with r = p/q, I can write f(x^{p/q}) = f(x^{p*(-q)}), at which point I can say that because both p and q are integers I have

f(x^{p*(-q)}) = (-p/q)f(x).

You mean, of course, f(x^p/(-q)) but where did the "-" come from? It's not necessary here. And why did you switch from log to f?

Oh wait. I've just realized that f(x^{-n}) \neq (-n)f(x). The negative sign shouldn't be there: I should be trying to get

f(x^{-n}) = (1/n)f(x).

Now I'm completely lost. I think I went wrong with the true statement
nlog(a) = (-m)log(a), because I forgot that (-m) is positive. After this error I can't seem to get back on track. Help?

Don't use both m and n: you mean m+ n= 0 so that m= -n. Just use n and -n.

EDIT: Looking at it again, I've realized another mistake:
f(x^{p*(-q)}) = pf(x) - qf(x), not what I stated before.

 

[brh2113]

I was mistakenly thinking of 1/x^{-1} = x in order to write p/q as p*(-q), but now I see that it should be p*q^{-1}, which ruins my entire plan.

So backtracking, I think now I've reduce the problem to proving that f(x^{1/q}) = (1/q)f(x), because I already know that I can bring the p down from before so I can ignore it for the moment while I true to prove this property for 1/q. Here I'm stuck, though.

I switched to f(x) because that's how we're supposed to write the problem, and it didn't dawn on me that I should be writing all of the steps of the proof that way until I was half way through.

 

[brh2113]

I've thought about your suggestions to go back and prove that

log(a^{nx}) = n log(a^{x}) for x as any real number.

If x=(1/n), then log(a^{nx}) = n log(a^{1/n}) = log (a).


log(a) is the same as n(1/n)log(a), but I'm not sure if this proves that

the 1/n can be brought down or if it just shows that in this case such happens to

be the case. Should it prove that that is the case, though, then I will have shown that 1/n

can be treated with the power rule, in which case I can say that r = p/q can

work with the power rule, which is what I want to prove.