If a(a-7b)= -b^2 then prove that log((a+b)/3)=(loga+logb)/2)

Question

How do I do this?

Answer 1

First

Start with the polynomial. Multiply out and get this is standard form.

a^2-7ab= -(b^2) ----> a^2-7ab+b^2=0

 

This is kind of useless unless you find a way to factor things out. Remember: if it's in this form on and you need to do things with it, then try to factor any way you can.

What we want to do is find something useful. I'm looking for either (a+b)*(a+b) or (a+b)*(a-b) usually. 

 

What happens when I put the equation like this:

a^2+2ab+b^2-9ab=0 ------> a^2+2ab+b^2= 9ab

???

Factor the left side! ----> (a+b)^2= 9ab

Then (a+b)= 3*((ab)^1/2)

That is 

(a+b)/3=(ab)^1/2

 

Is this enough to get you through to the log part?

Knowing the properties of logs you can find a proof of the left side of the log formula that is exactly equal to ((log(a)+log(b))/2