I am getting A as the answer.
Question is 3^x+2/9 = 3^x*9/9 = 3^x
From Option A
9^x>1 means 3^2x>1..
(3^x)2>1 . so 3^x has to be positive as it can never be negative for positive values or negative values of x.
so 3^x is always greater than 1
So we can answer this question.
But with statement B.. if x>0 then it can be 0.00001 also. so 3^0.0001 is less than 1 and with higher value like 2 we get value greater than 1. so not possible to answer with B. So answer is A
hey akalyan try taking log with base 3 to the question and you will easily reach at Is x>0 ?
Note everywhere in this answer wherever I write log it means log with base 3
Taking log both sides we get
log(3^(x+2)/9) > log 1
---> log(3^x+2) - log9>0
---->x+2 -2 >0