# Proof Power Rule Logarithms

1. Oct 28, 2007

### 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?

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2. Oct 28, 2007

### 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

3. Oct 29, 2007

### HallsofIvy

Staff Emeritus
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(ar)= 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(am+n)= log(an)+ log(am). With m+n= 0, what does that tell you.

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

4. Oct 30, 2007

### 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.

Last edited: Oct 30, 2007
5. Oct 30, 2007

### HallsofIvy

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

Yes.

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

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

Last edited: Oct 30, 2007
6. Oct 30, 2007

### 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.

7. Oct 30, 2007

### brh2113

$$log(a^{nx})$$ = $$n log(a^{x})$$ 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.