Question

$x^2-px+\log 2\log 3=0$ has one root as $\log 2$. Find the value of $p$.

• A. $\log 6$
• B. $\log 3$
• C. $\log 5$
• D. $\log 1$

Answer 1

The correct option is A $\log 6$

$f\left(x\right)=x^2-px+log2log3$ has a root as $\log 2$

$\therefore f\left(log2\right)=0$

${\left(log2\right)}^2-\left(log2\right)p+log2log3=0$

$\left(log2\right)p=log2(log2+log3)$

$p=log2+log3$

$p=log6$ as $(\log a+\log b=\log ab)$