log3x+7(2a2+3)<0,∀aϵR, if x lies in the interval (-a,-b) then 3a+ b is ___
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Solution
By definition of logax, we must have 2a2+3>0 and 3x+7>0
1st is always true and 2nd ⇒x>−73
Now log3x+7(2a2+3)<0 ⇒log(2a2+3)log(3x+7)<0
Since 2a2+3>1 ∴log(2a2+3)>log1=0 i.e. +ive
Hence log(3x+7) must be - ive ∴3x+7<1⇒x<−2⇒xϵ(−73,−2)