i. If a^2 + b^2 = 3ab, show that log (a+b/√5) =1/2(log a + log b).

Question

i. If a²  + b²  = 3ab, show that  log \(\left(\cfrac{a+b}{\sqrt{5}}\right)=\cfrac{1}{2}\)(log a + log b).

ii. If a² + b² = 7ab, prove that 2 log \(\left(\cfrac{a+b}{3}\right)\) = log a + log b.

iii. If a² - 12ab + 4b² = 0, prove that log (a + 2b) = \(\cfrac{1}{2}\)(loga + logb) + 2 log 2.

Answer 1

i. a² + b² = 3ab

\(\therefore\) a² + 2ab + b² = 3ab + 2ab

(a + b)² = 5ab

ii. a² + b² = 7ab

\(\therefore\) a² + 2ab + b² = 7ab + 2ab

\(\therefore\) (a + b)² = 9ab

iii. a² - 12ab + 4b² = 0

\(\therefore\) a² + 4b² = 12ab

\(\therefore\) a² + 4ab + 4b² = 12ab + 4ab

\(\therefore\) (a + 2b)² = 16ab

Taking log on both sides, we get

log (a + 2b)² = log (16ab)

\(\therefore\) log(a + 2b)² = log16 + log a + log b

   ….[By product law]

\(\therefore\) log (a + 2b)² = (log a + log b) + log 2⁴

\(\therefore\) 2 log (a + 2b) = (log a + log b) + 4 log 2

    ….[By exponent law]

Dividing throughout by 2, we get

log (a + 2b) = \(\cfrac{1}{2}\) (log a + log b) + 2 log 2