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Question

Sum of all integral values of x satisfying the inequality (35/2log3(123x))(3log3x)>32 is

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Solution

(352log3(123x))(3log3x)>32 is valid only when
123x>0x<4 ... (i)
Also, x>0 ... (ii)
Now, 352log3(123x)3log3x>32
3log3(123x)523log3x>32

{(123x)52}x>32{alogab=b}
(123x)5232>x
(123x)52>x+32
(123x)5>(x+32)2(iii)

from (i),(ii),
xϵ(0,4)
there are 3 integral values in this range, we will check them with equation (iii)
So, we have to check for x=1,2,3
x=1[95>(33)29×812>332] True
x=2[65>(34)26×362>342] True
x=3[35>(35)2243>1125] False
x=1,2
Sum of integral values of x satisfying given equation is 1+2=3.

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