Question

If $\frac{{\log }_{2}a}{4}=\frac{{\log }_{2}b}{6}=\frac{{\log }_{2}c}{3k}$ and $a^3{b}^{2}c=1$, then the value of $k$ will be

• A. $-7$
• B. $-1$
• C. $-4$
• D. $-8$

Answer 1

The correct option is D $-8$

$a^3{b}^{2}c=1$

Taking log on both sides to the base 2, we get

${\log }_2\left(a^3{b}^{2}c\right)={\log }_{2}1$

$\Rightarrow {\log }_2\left(a^3\right)+{\log }_2\left(b^2\right)+{\log }_{2}c=0$

$\therefore 3{\log }_{2}a+2{\log }_{2}b+{\log }_{2}c=0$

$\therefore (\frac{4}{k}+\frac{4}{k}+1)\left({\log }_{2}c\right)=0$

Since ${\log }_{2}c\ne 0,\therefore \frac{8}{k}=-1$

$\therefore k=-8$