SOLUTION: if a^2+b^2=7ab,prove that log 1/3(a+b)=1/2[log a+log b]

Question 313421: if a^2+b^2=7ab,prove that log 1/3(a+b)=1/2[log a+log b]
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
if a^2+b^2=7ab,prove that log 1/3(a+b)=1/2[log a+log b]
log rule: logb (mn) = logb (m) + logb (n)
log rule: logb (m^n) = nlogb (m)
log [(1/3)(a + b)] = (1/2)(log a + log b)
log [(1/3)(a + b)] = (1/2)(log ab)
2log [(1/3)(a + b)] = log ab (multiplied both sides by 2)
log [(1/3)^2 * (a + b)^2] = log ab
log [(1/9)(a + b)^2] = log ab
(1/9)(a + b)^2 = ab (removed the logs)
(7/9)(a + b)^2 = 7ab (multiplied both sides by 7)
(7/9)(a^2 + 2ab + b^2) = 7ab (multiplied the left side out using FOIL)
and a^2+b^2 = 7ab (this is supposed to be true)
a^2 + b^2 = (7/9)(a^2 + 2ab + b^2) (set them equal)
9a^2 + 9b^2 = 7a^2 + 14ab + 7b^2 (multiplied both sides by 9)
2a^2 + 2b^2 = 14ab (subtracted 7a^2 + 7b^2 from both sides)
a^2 + b^2 = 7ab (divided both sides by 2)

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