Question

If $a=lo{g}_{24}12,b=lo{g}_{36}24,c=lo{g}_{48}36$, then $1+abc$ is equal to

• A. $2ac$
• B. $2bc$
• C. $2ab$
• D. None of these

Answer 1

The correct option is B

$2bc$

Using identity, $lo{g}_{b}a=\frac{lo{g}_{m}a}{lo{g}_{m}b}$

Also, $loga$ means $lo{g}_{10}a$ i.e. it has a base 10

$abc=\frac{log12}{log24}.\frac{log24}{log36}.\frac{log36}{log48}=\frac{log12}{log48}$

$\therefore 1+abc=\frac{log48+log12}{log48}=\frac{log\left(48.12\right)}{log48}$

$=\frac{log{24}^2}{log48}=2.\frac{log24}{log48}$ ($\because lo{g}_b{a}^m=m.lo{g}_{b}a$)

$=2.\frac{lo{g}_{36}24}{lo{g}_{36}48}=2bc$