Proving Logarithmic Rule: log_a(x^k)=klog_ax

[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

 

[cristo]

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).

 

[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

 

[fungfat]

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

You want to prove: log_a x^k = k log_a x

Proof:

(1) First, let y = log_a x

(2) Rewite (1) in exponential form using anti-log: a^y = x

(3) Next, add exponent k to both side and we have: a^yk = x^k

(4) Now you "log" both side to have: log_a a^yk = log_a x^k

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

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

(6) so (4) becomes: yk = log_a x^k

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

(8) so (6) becomes (log_a x ) (k) = log_a x^k

(9) Re-arrange the left, we have: k log_a x = log_a x^k

(10) Proof complete QED