Is 3^(x+2)/9 1 ? (1) 9^x 1 (2) x 0

Question

Is \frac{3^{(x+2)}}{9} >1 ? (1) 9^x > 1 (2) x > 0 I solved it. For example, in clue (1), I did this: 9^x > 1 9^x > 9^0 Then, x > 0 , SUFFICIENT. However, I would like to know whether there is a more conceptual approach. Specifically, whether in which cases x can be positive or negative. Source: https://www.gmathacks.com

Bunuel · Answer

Is 3^(x+2)/9 >1 ?

Is  \frac{3^{x+2}}{9} >1 ? --> is  \frac{3^{x+2}}{3^2} >1 ? --> is  3^x>1 ? --> is  x>0 ?

Or: is  \frac{3^{x+2}}{9} >1 ? --> is  3^{x+2}>3^2 ? --> is  x+2>2 ? --> is  x>0 ?

(1) 9^x > 1 -->  x>0 . Sufficient.
(2) x > 0. Directly answers the question. Sufficient.

Answer: D.

Hope it's clear.

akalyan · Answer

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

With

Bunuel · Answer

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 . For more on number theory and exponents check: math-number-theory-88376.html DS questions on exponents: search.php?search_id=tag&tag_id=39 PS questions on exponents: search.php?search_id=tag&tag_id=60 Tough and tricky DS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125967.html Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html Hope it helps.

manimani · Answer

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

akalyan · Answer

Thanks for correcting me bunuel.

Yes taking log is also good way to solve this problem manimani.

metallicafan · Answer

Hi Bunuel,

I would like to know whether you have a link with theory about exponents with inequalities.

For example, we cannot say inmediately that because  a^x > a^y  , then  x > y , right?

Because  a  could be a fraction.

Please, if you have theory about that specific point, please post it.
Thanks!

fmomen01 · Answer

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....

aepmk · Answer

Is the answer A or B ?

Sent from my iPhone using GMAT Club Forum

Bunuel · Answer

You can check the OA of a question under the spoiler in the first post. The OA for this question is D. Solutions are given above.

Is 3^(x+2)/9 >1 ? (1) 9^x > 1 (2) x > 0

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