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Question

Consider the four sets A,B,C and D defined as
A={a:a=8(log4(x))3+4log2(x4), x>0}
B={b:b=20+13(logx2)2, x>0}
C={cN:cAB for same x>0}
D={xZ:(x2)2(15x256x+17)<0}

Then the number of elements in C×D is

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Solution

C denotes the common elements (natural numbers) in A and B.
8(log4(x))3+4log2(x4)=20+13(logx2)2
8(log4x)3+4log2(x4)2013(log2x)2=0 ; x1
8×(12)3(log2x)3+4×2×4log2x13(log2x)220=0
(log2x)313(log2x)2+32log2x20=0

Let log2x=t
Then t313t2+32t20=0
By trial and error method, we find that t=1 is a root.

Applying synthetic division
t=11133220011220112200

t313t2+32t20=(t1)(t212t+20)=0
(t1)(t2)(t10)=0
t=1,2,10
x=21,22,210=2,4,1024
C={2,4,1024}

(x2)2(15x256x+17)<0
15x256x+17<0 ;x2
15x25x51x+17<0
(3x1)(5x17)<0
(x13)(x175)<0
x(13,175){2}
D={1,3}

n(C×D)=n(C)×n(D)=3×2=6

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