Question

If $a={\log }_x\left(yz\right)$, $b={\log }_y\left(zx\right)$, $c={\log }_z\left(xy\right)$, then by symmetry $a,b,c$ are equal and their common value is

• A. $2$
• B. $3$
• C. $4$
• D. $7$

Answer 1

Answer: (A)

$2$

Given ${\log }_x\left(yz\right)=a$

$\Rightarrow {\log }_{x}x+{\log }_x\left(yz\right)={\log }_{x}x+a$

$\Rightarrow {\log }_x\left(xyz\right)=1+a$

$\Rightarrow {\log }_{xyz}x=\frac{1}{1+a}$

Similarly we get

${\log }_{xyz}y=\frac{1}{1+b}$ and ${\log }_{xyz}z=\frac{1}{1+c}$

Adding the three equations we get,

${\log }_{xyz}xyz=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$

$\Rightarrow 1=\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$

Since $a,b$ and $c$ are equal we get

$\frac{3}{1+a}=1\Rightarrow a=2$

Hence, $a=b=c=2$