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Question

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 loga x, 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)>log 1=0 i.e. +ive
Hence log(3x+7) must be - ive
3x+7<1x<2xϵ(73,2)

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