Question

If $\frac{{\log }_{1/2}x}{b-c}=\frac{{\log }_{1/2}y}{c-a}=\frac{{\log }_{1/2}z}{a-b}$ , then the value of $x^a{y}^b{z}^c$ is equal to

• A. $\frac{1}{2}$
• B. $xyz$
• C. ${\left(\frac{1}{2}\right)}^{abc}$
• D. $1$

Answer 1

The correct option is D $1$

Let $\frac{{\log }_{1/2}x}{b-c}=\frac{{\log }_{1/2}y}{c-a}=\frac{{\log }_{1/2}z}{a-b}=k$
Now, $a{\log }_{1/2}x+b{\log }_{1/2}y+c{\log }_{1/2}z=k[a(b-c)+b(c-a)+c(a-b)]$
$\Rightarrow a{\log }_{1/2}x+b{\log }_{1/2}y+c{\log }_{1/2}z=0$
$\Rightarrow {\log }_{1/2}{x}^a{y}^b{z}^c=0[\because a\log x=\log x^a\text{\&}\log a+\log b+\log c=\log (abc\left)\right]$
$\Rightarrow x^a{y}^b{z}^c=1$
Ans: D