Question

If 1/4 log a base 2 = 1/6 log b base 2 = 1/24 log c base 2 the value of a^3 b^2 c is ?

Please Give Full Solution

(b) 1 is Correct Answer ​

Answer 1

Given,

$\longrightarrow\sf{\dfrac{1}{4}\log_2a=\dfrac{1}{6}\log_2b=-\dfrac{1}{24}\log_2c}$

Consider,

$\longrightarrow\sf{\dfrac{1}{4}\log_2a=-\dfrac{1}{24}\log_2c}$

$\longrightarrow\sf{\dfrac{\log_2a}{4}=\dfrac{-\log_2c}{24}}$

Since $\sf{b\log a=\log\left(a^b\right),}$

$\longrightarrow\sf{\dfrac{\log_2a}{4}=\dfrac{\log_2(c^{-1})}{24}}$

By cross multiplication,

$\longrightarrow\sf{24\log_2a=4\log_2(c^{-1})}$

Dividing by 4,

$\longrightarrow\sf{6\log_2a=\log_2(c^{-1})}$

$\longrightarrow\sf{\log_2(a^6)=\log_2\left(\dfrac{1}{c}\right)}$

Taking antilog,

$\longrightarrow\sf{a^6=\dfrac{1}{c}\quad\quad\dots(1)}$

Consider,

$\longrightarrow\sf{\dfrac{1}{6}\log_2b=-\dfrac{1}{24}\log_2c}$

$\longrightarrow\sf{\dfrac{\log_2b}{6}=\dfrac{-\log_2c}{24}}$

$\longrightarrow\sf{\dfrac{\log_2b}{6}=\dfrac{\log_2(c^{-1})}{24}}$

By cross multiplication,

$\longrightarrow\sf{24\log_2b=6\log_2(c^{-1})}$

Dividing by 6,

$\longrightarrow\sf{4\log_2b=\log_2(c^{-1})}$

$\longrightarrow\sf{\log_2(b^4)=\log_2\left(\dfrac{1}{c}\right)}$

Taking antilogs,

$\longrightarrow\sf{b^4=\dfrac{1}{c}\quad\quad\dots(2)}$

Multiplying (1) and (2),

$\longrightarrow\sf{a^6b^4=\dfrac{1}{c^2}}$

$\longrightarrow\sf{a^6b^4c^2=1}$

$\longrightarrow\sf{(a^3b^2c)^2=1}$

Since logarithm is defined only for positive integers,

$\sf{\longrightarrow\underline{\underline{a^3b^2c=1}}}$

Hence (b) is the answer.