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Question

Find the solution set in log2(x1)+log2(x2)>1


A

x>1

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B

x>2

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C

x>3

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D

(,0) U (3,)

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Solution

The correct option is C

x>3


log2(x1)+log2(x2)>1

For Logarithm to be defined

x - 1 > 0, x > 1 ------------------(1)

And x - 2 > 0, x > 2 ------------(2)

This inequality can also be written as

log2(x1)(x2) > 1

log2(x23x+2) > 1

Since, base of the logarithm greater than 1. So, given logarithm is a decreasing function.

x23x+2>21

x23x+22>0

x23x>0

(x)(x3)>0 x(,0)(3,) ---(3)

So,

From equation 1,2 and 3 Or common part of equation 1,2 and 3

We get,

x > 3


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