Question

Let $a=\left({\log }_3\pi \right)\left({\log }_{2}3\right)\left({\log }_{\pi }2\right)$,
$b=\frac{\log 576}{3\log 2+\log 3}$ the base of the logarithm being 10 , $c=2($ sum of the solution of the equation $(3{)}^{4x}-(3{)}^{\left(2x+{\log }_3(12)\right)}+27=0)$ and $d=7^{\left({\log }_{7}2+{\log }_{7}3\right)}$ then $(a+b+c÷d)$ simplifies to

• A. rational which is not natural.
• B. natural but not prime.
• C. irrational
• D. even but not composite

Answer 1

Answer: (A)

$a=\left({\log }_3\pi \right)\left({\log }_{2}3\right)\left({\log }_{\pi }2\right)=\frac{\log \pi }{\log 3}\cdot \frac{\log 3}{\log 2}\cdot \frac{\log 2}{\log \pi }=1$
$b=\frac{\log 576}{3\log 2+\log 3}\text{ the base of the log being }10=\frac{2\log 24}{\log 24}=2$
$c=(9{)}^{2x}-9^x\cdot 3^{{\log }_{3}12}+27=0\Rightarrow t^2-12t+27=0$
$\Rightarrow t=9\Rightarrow 9^x=9\Rightarrow x=1$
$\Rightarrow t=3\Rightarrow 9^x=3\Rightarrow x=\frac{1}{2}$
$\Rightarrow c=2\left(1+\frac{1}{2}\right)=3$
$d=7^{\left({\log }_{7}2+{\log }_{7}3\right)}=6$
$a+b+\frac{c}{d}=1+2+\frac{3}{6}=\frac{7}{2}$