Question

a = log {12 base 24}, b = log {24 base 36}, c = log {36 base 48}, then 1 + abc is equal to?​

Answer 1

Given :-

$\rm \rightarrow a = log_{24}(12) \\ \rm\rightarrow b = log_{36}(24) \\\rm\rightarrow c = log_{48}(36)$

To find :-

$\rm 1+abc$

Solution :-

$\rm \longrightarrow 1+abc$

$\rm\longrightarrow1 + \bigg( log_{24}(12) \times log_{36}(24) \times log_{48}(36) \bigg)$

$\rm\longrightarrow1 + \bigg( \dfrac{log(12)}{log(24)} \times \dfrac{log(24)}{log({36})} \times \dfrac{log(36)}{log({48})} \bigg) \\ \\ \\ \rm\longrightarrow1 + \bigg( \dfrac{log(12)}{1} \times \dfrac{1}{1} \times \dfrac{1}{log({48})} \bigg)\\ \\ \\ \rm\longrightarrow1 + \bigg( \dfrac{log(12)}{log({48})} \bigg)\\ \\ \\ \rm\longrightarrow 1 + log_{48}(12) \\ \\ \\ \rm\longrightarrow log_{48}(48) + log_{48}(12) \\ \\ \\ \rm\longrightarrow log_{48 }(48 \times 12) \\ \\ \\ \rm\longrightarrow log_{48}{576}\\ \\ \\ \rm\longrightarrow log_{48}{ {(24)}^{2} } = 2log_{48}{ {(24)} }\\ \\ \\ \rm\longrightarrow 2log_{48}{ {(24)} } \times log_{36}{ {(36)} }\\ \\ \\ \rm\longrightarrow 2 \times \dfrac{log_{24}}{log_{48}} \times \dfrac{log_{36}}{log_{36}} \\ \\ \\ \rm\longrightarrow 2 \times \dfrac{log_{24}}{log_{36}} \times \dfrac{log_{36}}{log_{48}} \\ \\ \\ \rm\longrightarrow 2 \times b \times c\\ \\ \\ \rm\longrightarrow 2bc$

Answer :-

The correct answer is option (c) 2bc

Answer 2

Step-by-step explanation:

the correct answer is c 2bc