# Thread: prove (log a b^2).(log b a^3) = 6

1. ## prove (log a b^2).(log b a^3) = 6

can anyone help with this question? cheers

2. Originally Posted by 24680
can anyone help with this question? cheers
Before I start I would like to suggest that you repeat the problem statement in the written portion of your post. It would be easier for us to see what you are trying to write, particularly when we get LaTeX back.

I presume that "log a b^2" is "log to the base a of b^2? Let me rewrite that as log(a)b^2.

[log(a)b^2]*[log(b)a^3] = [2*log(a)b]*[3*log(b)a] = 6*[log(a)b]*[log(b)a]

There are probably several ways to finish this. I am going to use the change of base formula to change both of these to log base 10 (which I will write simply as "log x.")

The change of base formula is:
log(k)x = (log x)/(log k) to convert to log base 10.

So
log(a)b = (log b)/(log a)
log(b)a = (log a)/(log b)

So
[log(a)b^2]*[log(b)a^3] = 6*[log(a)b]*[log(b)a] = 6*(log b)/(log a)*(log a)/(log b) = 6

-Dan

3. Originally Posted by 24680
can anyone help with this question? cheers
(log_a b^2).(log_b a^3)

..................= [(log_b b^2)/(log_b (a))].(log b a^3)

..................= (log_b b^2).[(log b a^3)/(log_b (a))]

..................= (log_b b^2).(log a a^3)

..................= 2x3 = 6

RonL

4. Originally Posted by 24680
can anyone help with this question? cheers
Hello,

you only have to use the property: a^(log_a(b)) = b

expand the powers:

2*log_a(b) * 3 * log_b(a) = 6.

6 * [log_a(b) * log_b(a)] = 6

6 * [log_a(b^(log_b(a)))] = 6

6 * [log_a(a)] = 6

6 = 6

tsch&#252;ss

EB