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

• Oct 8th 2006, 11:55 AM
24680
prove (log a b^2).(log b a^3) = 6
can anyone help with this question? cheers
• Oct 8th 2006, 12:26 PM
topsquark
Quote:

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
• Oct 8th 2006, 12:29 PM
CaptainBlack
Quote:

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
• Oct 8th 2006, 12:34 PM
earboth
Quote:

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