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Question

Solve the following inequality:
log0.5(4x)log0.52log0.5(x1)

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Solution

log2(4x)log22+log2(x1)
log22log2(4x)log2(x1)
Taking antilog,we get,
24xx1
2(4x)(x1)4x0
2(4x4x2+x)4x0
25x+x2+4x40
x25x+6x40
(x3)(x2)(x4)0therefore,
xε(,2][3,4)
Now,asx<4andx>1
xε(1,4)
so,wecansaythat,
xε(1,2][3,4)

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