Logarithmic Query

[BigGlenntheHeavy]

\(\displaystyle If \ a^2+b^2 \ = \ 7ab, \ show \ that \ log\bigg(\frac{|a+b|}{3}\bigg) \ = \ \frac{1}{2}(log|a|+log| b|).\)

 

[Deleted member 4993]

\(\displaystyle (a + b)^2 \ \ = \ \ 9ab\)

\(\displaystyle \left (\frac{a + b}{3}\right )^2 \ \ = \ \ ab\)

 

[wjm11]

If a^2 + b^2 = 7ab, show that log(|a + b|/3) = (1/2)(log|a| + log|b|).

If a^2 + b^2 = 7ab, then a and b must both be positive or both be negative.
Additionally, if log(|a + b|/3) = (1/2)(log|a| + log|b|) can be evaluated, then neither a nor b is equal to 0.

Going the "other" direction...

log(|a + b|/3) = (1/2)(log|a| + log|b|) Multiply both sides by 2.
log ([(a + b)^2]/9) = (log|a| + log|b|) Expand left side; consolidate right side.
log ([a^2 + 2ab + b^2]/9) = log(ab)
[a^2 + 2ab + b^2]/9 = ab Multiply both sides by 9.
a^2 + 2ab + b^2 = 9ab Subtract 2ab from both sides.
a^2 + b^2 = 7ab

 

[BigGlenntheHeavy]

Right on wmj11, the easiest way to solve this problem, is to go trhrough "the back door".