Question

If sum of the roots of ${\log }_2(3^{2+x}-6^x)=3+x{\log }_2\left(\frac{3}{2}\right)$ is $P$, then which of the following is/are correct?

• A. $P$ is prime number
• B. ${\log }_{3}P+{\log }_{P}3=2$
• C. ${\log }_3(P-1)=0$
• D. ${\log }_3(P+1)=0$

Answer 1

Answer: (A), (B)

The correct options are
A $P$ is prime number
B ${\log }_{3}P+{\log }_{P}3=2$

Given : ${\log }_2(3^{2+x}-6^x)=3+x{\log }_2\left(\frac{3}{2}\right)$
$\Rightarrow {\log }_2\left(\frac{3^{2+x}-6^x}{{\left(\frac{3}{2}\right)}^x}\right)=3\Rightarrow \frac{2^x\cdot 3^x(9-2^x)}{3^x}=8\Rightarrow 2^x(9-2^x)=8$
Assuming $2^x=t$, we get
$\Rightarrow t(9-t)=8\Rightarrow t^2-9t+8=0\Rightarrow (t-8)(t-1)=0\Rightarrow t=1,8\Rightarrow 2^x=1,8\Rightarrow x=0,3\Rightarrow P=0+3=3$

Check the options.