# Log proof

1. May 20, 2007

### Hypochondriac

prove:
$$log_a(x^k)=klog_ax$$

i can do the addition and subtraction rules no problem, but for some reason i'm stuggling with this one,
just point me in the right direction, or help with the whole thing i'm not bothered, i just hope it aint in my exam tomorrow lol

2. May 20, 2007

### cristo

Staff Emeritus
Well, one way of doing it is to write log(x^k) as log(x.x.x...x) [where there are k x's in the argument of the logarithm]. Then invoke the rule log(ab)=log(a)+log(b). Of course this only works for positive integers!

A better proof is to let y=log_a(x), so that a^y=x. Then x^k=(a^y)^k=a^(yk). Hence, log_a(x^k)=ky=k log_a(x).

3. May 20, 2007

### Hypochondriac

thanks
but to get from x^k = a^(yk)
to log_a(x^k) = ky

you would need to go via the rule i'm trying to prove
ie log_a(x^k) = log_a(a^ky) = ky log_a(a) = 1ky
so i didnt know if that was a feasible move

4. Feb 17, 2009

### fungfat

No, you do not have to use the log rule that you are trying to prove to continue...

You want to prove: loga xk = k loga x

Proof:

(1) First, let y = loga x

(2) Rewite (1) in exponential form using anti-log: ay = x

(3) Next, add exponent k to both side and we have: ayk = xk

(4) Now you "log" both side to have: loga ayk = loga xk

(by logging both side you do not use the log rule you are trying to prove)​

(5) but we know the log rule said: logaayk = yk

(6) so (4) becomes: yk = loga xk

(7) but (1) said y = loga x

(8) so (6) becomes (loga x ) (k) = loga xk

(9) Re-arrange the left, we have: k loga x = loga xk

(10) Proof complete QED