Question

# Sum of all integral values of x satisfying the inequality  352log3(12−3x)−3log2x>32 is

Solution

## (32.5log3(12−3x))−(3log2x)>32 ⇒12−3x>0 and x>0 ⇒x∈(0,4) Therefore, the integral values are 1,2,3. For x=1, (32.5log39)−(3log21)=35−30>32 For x=2, (32.5log36)−(3log22)=(65)1/2−31>32 We know that, 62−3>32∴(65)1/2−31>32 For x=3, (32.5log33)−(3log23)=32.5−3log23<32 We know that, 33=27<32∴32.5−3log23<32 Hence, the required values are 1,2 and the required sum =3

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