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I have no idea on how to start solving this question:

Given that p^2+q^2=11pq, where p and q are constants, show that 1/2(logp+logq) equals:
(a)log(p-q/3)
(b)log(p+q/surd13)
Could you solve for me?

Questioner:JohnnyX
Country:California, United States
Private:No
Subject:Logarithmic Proof

Question:
I have no idea on how to start solving this question:

Given that p^2+q^2=11pq, where p and q are constants, show that 1/2(logp+logq) equals:
(a)log(p-q/3)
(b)log(p+q/surd13)
Could you solve for me?
...................................
Your equation, p^2+q^2=11pq,  looks mighty suspicious.  It makes me think that it was supposed to say

p^2+q^2 = 2pq, which would mean
p^2 - 2pq + q^2 = 0;
(p - q)^2 = 0

Or, maybe

p^2+q^2 = -2pq, which would mean
p^2 + 2pq + q^2 = 0,
(p + q)^2 = 0

But it doesn't.  It says

p^2+q^2 = 11 pq = 13pq - 2pq ,
(p + q)^2 = 13pq

So p + q = sqrt(13pq)

Now let's do the log stuff:

log(p + q) = log sqrt(13pq)

log(p + q) = 1/2 log (13pq)

log(p + q) = 1/2 [log (13) + log p + log q)]

log(p + q) = log sqrt(13) + 1/2[log p + log q)

You can finish up.

By the way, referring to sqrt(13) as 'surd(13)' was already obsolete before your grandfather was born.

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