Question

Positive numbers $x,y$ and $z$ satisfy
$xyz={10}^{81}$ and $\left({\log }_{10}x\right)\left({\log }_{10}yz\right)+\left({\log }_{10}y\right)\left({\log }_{10}z\right)=468.$ Then the value of $({\log }_{10}x{)}^2+({\log }_{10}y{)}^2+({\log }_{10}z{)}^2$ is

• A. $5625$
• B. $5652$
• C. $5265$
• D. $5526$

Answer 1

The correct option is A $5625$

Let ${\log }_{10}x=a,{\log }_{10}y=b,{\log }_{10}z=c$
Given, $xyz={10}^{81}$
$\Rightarrow {\log }_{10}x+{\log }_{10}y+{\log }_{10}z=81$
$\Rightarrow a+b+c=81$
Also, $a(b+c)+bc=468$
$\therefore a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ca)$
$=(81{)}^2-2(468)=5625$