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Question

Sum of all integral values of x satisfying the inequality 
352log3(123x)3log2x>32 is 


Solution

(32.5log3(123x))(3log2x)>32 123x>0 and x>0
x(0,4)
Therefore, the integral values are 1,2,3.

For x=1,
(32.5log39)(3log21)=3530>32

For x=2,
(32.5log36)(3log22)=(65)1/231>32
We know that,
623>32(65)1/231>32

For x=3,
(32.5log33)(3log23)=32.53log23<32
We know that,
33=27<3232.53log23<32

Hence, the required values are 1,2 and the required sum =3

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