Solve the inequality \(\displaystyle{x}^{{{\left({{\log}_{{2}}{x}}\right)}}}+{2}\leq{8}\)

ashes86047xhz

ashes86047xhz

Answered question

2022-03-20

Solve the inequality x(log2x)+28

Answer & Explanation

enriuadaziaa

enriuadaziaa

Beginner2022-03-21Added 7 answers

Taking logarithm wrt base 2,
(log2x+2)log2xlog28=3 
Write log2x=a
a2+2a30(a+3)(a1)03a1

Jambrichp2w2

Jambrichp2w2

Beginner2022-03-22Added 12 answers

The inequality is
x2(xlog2x)<8
Let x=2t. Then the inequality becomes
22t(2t)t<822t+t2<23
which is equivalent to 2t+t2<3.

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