Question

Number of prime numbers which are the factors of product of solution of the equation $|x-4{|}^{{\log }_2^{2}x+8}=x^{6{\log }_2(x-4)}$ is equal to

Answer 1

$|x-4{|}^{{\log }_2^{2}x+8}=x^{6{\log }_2(x-4)}$ where $(x-4)>0$

$\Rightarrow (x-4{)}^{({\log }_{2}x{)}^2+8}=(x-4{)}^{6{\log }_{2}x}$
Equating power of $(x-4)$
$\Rightarrow ({\log }_{2}x{)}^2+8=6{\log }_{2}x$ or $(x-4)=1$

$\Rightarrow ({\log }_{2}x{)}^2-6{\log }_{2}x+8=0\Rightarrow {\log }_{2}x=4,{\log }_{2}x=2\Rightarrow x=16,x=4$

$x=4$ not possible

when $x-4=1\Rightarrow x=5$

$\Rightarrow$ product of solutions

$=16\times 5=80=2^4\times 5$

number of prime numbers $=2$