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Question

The range of values of x that satisfy the inequation (4x+116x<2log48) is

A
x(0,log43)
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B
x(0,log34)
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C
x(0,log43]
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D
None of these
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Solution

The correct option is B x(0,log43)
Given,
(4x+116x<2log48)

Now,
log48=log423=3log42=3log24=32log22=32.............(1)

Let,
4x=t then given inequation reduce in the form

4tt2>2.32..........from(1)

t24t+3<0

(t1)(t3)<0

1<t<3

1<4x<3

0<x<log43

Hence, the answer is option A


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