Question

If $\log \left(\frac{a-b}{2}\right)=\frac{1}{2}(\log a+\log b)$, then the value of $a^2+b^2-6ab$ is equal to

• A. $3$
• B. $7$
• C. $0$
• D. $6$

Answer 1

Answer: (C)

$0$

$\log \left(\frac{a-b}{2}\right)=\frac{1}{2}(\log a+\log b)$

$\log \left(\frac{a-b}{2}\right)=\frac{1}{2}\log (a\times b)$ ...$\log (a\cdot b)=\log a+\log b$)

$\therefore \frac{a-b}{2}=a{b}^{\frac{1}{2}}$ ....removing log

Squaring both the sides we get

$\therefore \frac{(a-b{)}^2}{4}=ab$

$\therefore a^2+b^2-2ab=4ab$

$\therefore a^2+b^2-6ab=0$