Question

If $\frac{{\log }_{2}a}{4}=\frac{{\log }_{2}b}{6}=\frac{{\log }_{2}c}{3p}$ and also $a^3{b}^{2}c=1$, then the value of $p$ is equal to

• A. $-6$
• B. $-7$
• C. $-8$
• D. $-9$

Answer 1

Answer: (C)

$-8$

$\frac{{\log }_{2}a}{4}=\frac{{\log }_{2}b}{6}=\frac{{\log }_{2}c}{3p}=x{\log }_{2}a=4x$

$\Rightarrow a=2^{4x}$

${\log }_{2}b=6x$

$\Rightarrow b=2^{6x}$

${\log }_{2}c=3px$

$\Rightarrow c=2^{3px}$

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

$2^{12x}{\cdot 2}^{12x}\cdot 2^{3px}=2^0$

$2^{24x+3px}=2^0$

$\therefore 24x+3px=0$

$\therefore 3px=-24x$

$\therefore p=-8$