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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

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

With

The red part is not correct: \(3^{0.0001}\approx{1.000109867}>1\).

Generally \(3^x>1\) to hold true \(x\) must be more than zero, hence \(3^x>1\) simply means that \(x>0\).

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

With

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

Question is 3^(x+2)/9>1 Taking log both sides we get log(3^(x+2)/9) > log 1 ---> log(3^x+2) - log9>0 ---->x+2 -2 >0 ----> x>0
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This was so cool you guys! I got it wrong and at first I didn't understand any of the explanations, but once I followed them and wrote them down, I got it!! Thanks! ....As you can probably guess, I'm new here. I promise not to be so annoying, once I get the hang of this....

gmatclubot

Re: Is 3^(x+2)/9 > 1 ?
[#permalink]
22 Oct 2013, 04:45

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