Question

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

Answer 1

concepts : $log(AB)=logA+logB$.....(1)

$log\frac{A}{B}=logA-logB$........(2)

$logA^n=nlogA$.........(3)

it is given that, a² - 12ab + 4b² = 0

or, a² + 4b² = 12ab

or, a² + 4b² + 4ab = 4ab + 12ab

or, a² + 4b² + 4ab = 16ab

or, (a + 2b)² = 16ab

taking log both sides,

log(a + 2b)² = log(16ab)

or, 2log(a + 2b) = log(16ab) [ using formula (3), ]

or, 2log(a + 2b) = log16 + loga + logb [ using formula (1), ]

or, 2log(a + 2b) = log2⁴ + loga + logb

or, 2log(a + 2b) = 4log2 + loga + logb

or, log(a + 2b) = 2log2 + 1/2[ loga + logb ]

hence, $\log (a + 2b) = \frac{1}{2} (\log a + \log b) + 2\log 2$