Question

If,$\frac{{\log }_{2}x}{4}=\frac{{\log }_{2}y}{6}=\frac{{\log }_{2}z}{3k}$ and
$x^3.y^2.z=1$ than find value of K

• A. $8$
• B. $-8$
• C. $7$
• D. $10$

Answer 1

Answer: (B)

$-8$

Given $x^3{y}^{2}z=1$

$log\left(x^3{y}^{2}z\right)=log1$

$log{x}^3+log{y}^2+logz=log1$

$3logx+2logy+logz=0$

Given

$\frac{lo{g}_{2}x}{4}=\frac{lo{g}_{2}y}{6}=\frac{lo{g}_{2}z}{3k}$

$\frac{logx}{log2.4}=\frac{log4}{6log2}=\frac{logz}{3klog2}$

$\frac{logx}{4}=\frac{logy}{6}=\frac{logz}{3k}$

$logx=\frac{4}{3k}logz$ eqn(1)

$logy=\frac{6}{3k}logz$ eqn(2)

$3\left(\frac{4}{3k}logz\right)+2\left(\frac{6}{3k}logz\right)+logz=0$

$\log z.(\frac{4}{k}+\frac{4}{k}+1)=0$

$\frac{8}{k}+1=0$

$k=-8$