Question

If $a = \log_{24}12,b = \log_{36}24$and $c = \log_{48}36,$ then 1+abc is equal to

• A. $2ab$
• B. $2ac$
• C. $2bc$
• D. 0

Answer 1

Answer: (C)

$2bc$

Answer 2

$a = \log_{24}12 = \frac{\log 12}{\log 24} = \frac{2\log 2 + \log 3}{3\log 2 + \log 3}$

$b = \log_{36}24 = \frac{3\log 2 + \log 3}{2(\log 2 + \log 3)}$

$c = \log_{48}36 = \frac{2(\log 2 + \log 3)}{4\log 2 + \log 3}$

$\therefore abc = \frac{2\log 2 + \log 3}{4\log 2 + \log 3}$

⇒ $1 + abc = \frac{6\log 2 + 2\log 3}{4\log 2 + \log 3} = 2.\frac{3\log 2 + \log 3}{4\log 2 + \log 3} = 2bc$.