Proving Log Relationship: x2 + y2 = 11xy and log [(x-y)/3] = 0.5(logx + logy)

[TN17]

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.

 

[Mentallic]

Use these facts:

if $$\log(x)=\log(y)$$ then $$x=y$$

$$\log(x)+\log(y)=\log(xy)$$

and $$a\cdot \log(x)=\log(x^a)$$

 

[eumyang]

Given that x² + y² = 11xy

Show that log [ (x-y)/3 ] = 0.5 (logx +logy)

Here's a hint to start: subtract 2xy from both sides of
x² + y² = 11xy.