Question

Consider the four sets $A,B,C$ and $D$ defined as
$A=\{a:a=8({\log }_4\left(x\right){)}^3+4{\log }_{\sqrt{2}}\left(x^4\right),x>0\}$
$B=\{b:b=20+\frac{13}{({\log }_{x}2{)}^2},x>0\}$
$C=\{c\in N:c\in A\cap B\text{for same}x>0\}$
$D=\{x\in Z:(x-2{)}^2(15x^2-56x+17)<0\}$

Then the number of elements in $C\times D$ is

Answer 1

$C$ denotes the common elements (natural numbers) in $A$ and $B$.
$8\left({\log }_4\right(x){)}^3+4{\log }_{\sqrt{2}}(x^4)=20+\frac{13}{({\log }_{x}2{)}^2}$
$\Rightarrow 8({\log }_{4}x{)}^3+4{\log }_{\sqrt{2}}(x^4)-20-13({\log }_{2}x{)}^2=0;x\ne 1$
$\Rightarrow 8\times {\left(\frac{1}{2}\right)}^3({\log }_{2}x{)}^3+4\times 2\times 4{\log }_{2}x-13({\log }_{2}x{)}^2-20=0$
$\Rightarrow ({\log }_{2}x{)}^3-13({\log }_{2}x{)}^2+32{\log }_{2}x-20=0$

Let ${\log }_{2}x=t$
Then $t^3-13t^2+32t-20=0$
By trial and error method, we find that $t=1$ is a root.

Applying synthetic division
$\begin{array}{ccccc}t=1& 1& -13& 32& 20\\ & 0& 1& -12& -20\\ & 1& -12& -20& 0\end{array}$

$\Rightarrow t^3-13t^2+32t-20=(t-1)(t^2-12-20)=0$
$\Rightarrow (t-1)(t-2)(t-10)=0$
$\Rightarrow t=1,2,10$
$\Rightarrow x=2^1,2^2,2^{10}=2,4,1024$
$\therefore C=\{2,4,1024\}$

$(x-2{)}^2(15x^2-56x+17)<0$
$\Rightarrow 15x^2-56x+17<0;x\ne 2$
$\Rightarrow 15x^2-5x-51x+17<0$
$\Rightarrow (3x-1)(5x-17)<0$
$\Rightarrow (x-\frac{1}{3})(x-\frac{17}{5})<0$
$\Rightarrow x\in (\frac{1}{3},\frac{17}{5})-\left\{2\right\}$
$\therefore D=\{1,3\}$

$\therefore n(C\times D)=3\times 2=6$