# Thread: Prove this log using...

1. ## Prove this log using...

Given that x2 + y2 = 11xy
Show that log [ (x-y)/3 ] = 0.5 (logx +logy)

Similarly,
Give that log [ (x-y)/3 ] = 0.5 (logx +logy)
Show that x2 + y2 = 11xy
I really liked the log unit and understood it, but I was stumped on this question. It was for a test and it keeps bothering me.
What I did was completely wrong, but I will be able to understand it if someone shows the solution.

2. This is a good question. Here's a proof one way...hopefully it'll help you do it in reverse too.

We have x^2 + y^2 = 11xy.
Therefore x^2 - 11xy + y^2 = 0.

Note that (x - y)^2 = x^2 - 2xy + y^2.
Therefore (x - y)^2 = 9xy.

Take logs of both sides:
log((x - y)^2) = log(9xy).
Therefore 2log(x-y) = log(9) + log(x) + log(y).
(*) Therefore log(x-y) = 0.5(log(9) + log(x) + log(y)).

Now log((x-y)/3) = log(x-y) - log(3).
Therefore log(x-y) = log((x-y)/3) + log 3.

Putting this into (*):
log((x-y)/3) = -log(3) + 0.5(log(9) + log(x) + log(y)).
Therefore log((x-y)/3) = -log(3) + log(3) + 0.5(log(x) + log(y)).

Done!

Hope this helps.

3. Hello, TN17!

A variation of exambox's solution . . .

$\displaystyle \text{Given that: }\:x^2+ y^2 \:=\: 11xy$

$\displaystyle \text{Show that: }\:\log\left(\dfrac{x-y}{3}\right) \:=\:\frac{1}{2}(\log x + \log y)$

We have: .$\displaystyle x^2 + y^2 \:=\:11xy$

Subtract $\displaystyle 2xy$ from both sides: .$\displaystyle x^2 - 2xy + y^2 \:=\:9xy$

Then we have: .$\displaystyle (x-y)^2 \:=\:(3\sqrt{xy})^2$

Take the positive square root: .$\displaystyle x-y \;=\;3\sqrt{xy} \quad\Rightarrow\quad \dfrac{x-y}{3} \;=\;(xy)^{\frac{1}{2}}$

Take logs: .$\displaystyle \log\left(\dfrac{x-y}{3}\right) \;=\;\log(xy)^{\frac{1}{2}} \;=\;\frac{1}{2}\log(xy)$

Therefore: .$\displaystyle \log\left(\dfrac{x-y}{3}\right) \;=\;\frac{1}{2}(\log x + \log y)$

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### x^2 y^2=11xy

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