Question

If sum of the solutions of the equation ${\log }_2(3^{2+x}-6^x)=3+x{\log }_2\left(\frac{3}{2}\right)$ is $P$, then

• 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)

$P$ is prime number
${\log }_{3}P+{\log }_{P}3=2$

${\log }_2(3^{2+x}-6^x)=3+x{\log }_2\left(\frac{3}{2}\right)$
${\log }_2\left(\frac{3^{2+x}-6^x}{{\left(\frac{3}{2}\right)}^x}\right)=3[\because m\log x=\log x^m\text{\&}\log a-\log b=\log \frac{a}{b}]$
$\Rightarrow 2^3=\frac{3^{2+x}-6^x}{{\left(\frac{3}{2}\right)}^x}$
$\Rightarrow 8{\left(\frac{3}{2}\right)}^x=9\left(3^x\right)-3^x{2}^x$
$\Rightarrow (9-2^x)2^x=8$
Put $2^x=t$
$t^2-9t+8=0$
$(t-8)(t-1)=0$
$t=8,1$
$2^x=2^3,2^x=1$
$x=3,x=0$
Sum of solutions, $P=3$
Here, $P$ is a prime number.
So, option A is correct.
Now,${\log }_{3}P+{\log }_{p}3=1+1=2$
Option B is correct.
Also,${\log }_3(P-1)={\log }_{3}2\ne 0$
${\log }_3(P+1)={\log }_{3}4\ne 0$
Option C and D are incorrect