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Is 5^(x+2)/25<1 ? [#permalink]
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25 Jun 2012, 03:07
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Re: Is 5^(x+2)/25<1 ? [#permalink]
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25 Jun 2012, 03:07
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Re: Is 5^(x+2)/25<1 ? [#permalink]
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25 Jun 2012, 08:35
Hi,
Difficulty level: 600
To check, \(\frac {5^{(x+2)}}{25}<1\) or \((5^25^x)/25 < 1\) or \(5^x < 1\)?
Using (1), \(5^x<1\). Sufficient.
Using (2), x < 0 or \(5^x < 1\). Sufficient.
Answer (D)
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Re: Is 5^(x+2)/25<1 ? [#permalink]
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29 Jun 2012, 04:43



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Is 5^(x+2)/25<1 ? [#permalink]
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30 Jun 2015, 02:02
What would happen if we would have \(x=\frac{1}{10}\)? Am I correct that \(5^\frac{1}{10} = \frac{1}{5^(1/10)} = \frac{1}{\sqrt[10]{5^1}}\), but the denominator would still be always greater than the numerator? Thank you!
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Is 5^(x+2)/25<1 ? [#permalink]
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30 Jun 2015, 03:31
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bgpower wrote: What would happen if we would have \(x=\frac{1}{10}\)?
Am I correct that \(5^\frac{1}{10} = \frac{1}{5^(1/10)} = \frac{1}{\sqrt[10]{5^1}}\), but the denominator would still be always greater than the numerator?
Thank you! Hello bgpowerYou are correct that \(5^\frac{1}{10} = \frac{1}{5^(1/10)} = \frac{1}{\sqrt[10]{5^1}}\) But this statement "the denominator would still be always greater than the numerator?" is suspicious. If you talk about this fraction: \(\frac{1}{\sqrt[10]{5^1}}\) than you are wrong because in this fraction denominator \(\sqrt[10]{5^1}\) is less than nominator \(1\) If you talk about fraction from initial task then you are right denominator \(25\) is bigger than nominator \(\sqrt[10]{5^1}\)
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Is 5^(x+2)/25<1 ? [#permalink]
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30 Jun 2015, 03:42
Hi, Thanks for your reply! I was actually talking about: \(\frac{1}{\sqrt[10]{5^1}}\). So in this case I am wrong, as the numerator of 1 would clearly be greater than the denominator. But doesn't this mean that \(5^x\) is not always <1. The question does not define x to be positive and as shown above if x is a negative fraction (as \(\frac{1}{10}\)) then the result is greater than 1. Thanks for the clarification!
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Re: Is 5^(x+2)/25<1 ? [#permalink]
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30 Jun 2015, 04:39
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bgpower wrote: Hi, Thanks for your reply!
I was actually talking about: \(\frac{1}{\sqrt[10]{5^1}}\). So in this case I am wrong, as the numerator of 1 would clearly be greater than the denominator. But doesn't this mean that \(5^x\) is not always <1. The question does not define x to be positive and as shown above if x is a negative fraction (as \(\frac{1}{10}\)) then the result is greater than 1.
Thanks for the clarification! I think you leave out last step of task (dividing on 25) and this confuse you. You are absolutely right that \(\frac{1}{\sqrt[10]{5^1}}\) greater than 1 but if we divide this result on 25 (as tasks asks) then result will be less than 1  This task can be solved in much faster way: Tasks asks if \(\frac{5^{(x+2)}}{25}\) will be less than 1 Let's transform this equation to \(5^{(x+2)} < 25\) > \(5^{(x+2)} < 5^2\) from this view we see that this equation will be true if x will be less than 0 1) \(5^x<1\) this is possible only if \(x < 0\)  Sufficient 2) \(x<0\)  this is exactly what we seek  Sufficient Sometimes picking numbers is good but in this case algebraic way is much faster and at the end you will not have any hesitations in answer
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Is 5^(x+2)/25<1 ? [#permalink]
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30 Jun 2015, 05:10
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I actually did it the following way: \(\frac{5^(x+2)}{25}<1\) \(\frac{(5^x)(5^2)}{5^2}<1\) => Here you can basically cancel out \(5^2\) and are left with \(5^x<1\) Here we come to the point we have already discussed. (1) is clearly SUFFICIENT as it says exactly \(5^x<1\). (2) I thought (2) is NOT SUFFICIENT as for negative fractions (think \(\frac{1}{10}\)), IMO this does NOT hold true, while for other negative values it does.
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Re: Is 5^(x+2)/25<1 ? [#permalink]
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30 Jun 2015, 11:40
Is (5^x+2)/25<1 ? Multiply both sides by 25 to yield 5^x+2>25. Rewrite 25 so that we have 5 as a base on both sides, so we want to know if 5^x+2> 5^2. The question is now is x+2>2. This will be true if x is negative. (1) 5^x<1 1 can be rewritten so as 5^0 so as to give the same base. x<0 Sufficient. (2) x<0 This is the same information as we derived from Statement 1 (x is negative). Therefore it is also sufficient.
D



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Re: Is 5^(x+2)/25<1 ? [#permalink]
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30 Jun 2015, 11:41
Is (5^x+2)/25<1 ? Multiply both sides by 25 to yield 5^x+2<25. Rewrite 25 so that we have 5 as a base on both sides, so we want to know if 5^x+2< 5^2. The question is now is x+2<2. This will be true if x is negative. (1) 5^x<1 1 can be rewritten so as 5^0 so as to give the same base. x<0 Sufficient. (2) x<0 This is the same information as we derived from Statement 1 (x is negative). Therefore it is also sufficient.
D



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Is 5^(x+2)/25<1 ? [#permalink]
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01 Jul 2015, 03:56
OptimusPrepJanielleThank you for your explanation! I fully get your explanation. Nevertheless, I still don't see anyone addressing my questions, which basically asks what happens when x is a negative fraction as \(\frac{1}{10}\)? I may be the one with an error here, but please explain it to me. Thanks!
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Re: Is 5^(x+2)/25<1 ? [#permalink]
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01 Jul 2015, 04:09
Hi bgpower,
The issue is whether or not x is negative. If x is a negative fraction such as 1/10 the expression should still hold true. I hope that helps.



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Re: Is 5^(x+2)/25<1 ? [#permalink]
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01 Jul 2015, 04:12
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bgpower wrote: OptimusPrepJanielleThank you for your explanation! I fully get your explanation. Nevertheless, I still don't see anyone addressing my questions, which basically asks what happens when x is a negative fraction as \(\frac{1}{10}\)? I may be the one with an error here, but please explain it to me. Thanks! Hello bgpowerWhen x = 1/10 then \(5^{1/10+2} = 5^{19/10}\) We need to check whether this \(5^{19/10}\) less than \(5^2\) 19/10 less than 2 so \(5^{x+2} < 5^2\)
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Re: Is 5^(x+2)/25<1 ? [#permalink]
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21 Jul 2016, 14:18
For\frac{\(5^(x+2)\)}{25} < 1, the numerator will have to be less than 25.
\(5^(x+2)\) < 25?
For \(5^(x+2)\) < 25, x will have to be less than 0. x<0?
So statement B is straightforward. I am bungled up on the first statement.
\(5^x\) <1
Since \(5^0\) = 1, x != 0. Also, \(5^1\) = 5 and hence x<1.
But what about x being a fraction i.e. 0<x<1?
If that is the case then Statement 1 is no sufficient.
Please tell me where I am wrong.



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Is 5^(x+2)/25<1 ? [#permalink]
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21 Jul 2016, 14:41
Witchcrafts wrote: For\frac{\(5^(x+2)\)}{25} < 1, the numerator will have to be less than 25.
\(5^(x+2)\) < 25?
For \(5^(x+2)\) < 25, x will have to be less than 0. x<0?
So statement B is straightforward. I am bungled up on the first statement.
\(5^x\) <1
Since \(5^0\) = 1, x != 0. Also, \(5^1\) = 5 and hence x<1.
But what about x being a fraction i.e. 0<x<1?
If that is the case then Statement 1 is no sufficient.
Please tell me where I am wrong. Hello Witchcraftsif x is fraction than it will be some root from 5 and result of any root will be always more than 1 so x can't be a fraction if \(x = \frac{1}{2}\) then \(5^x = \sqrt{5}\)
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Re: Is 5^(x+2)/25<1 ? [#permalink]
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21 Jul 2016, 15:38
Thanks Harley. I figured that after I posted the question The rule is: If a x and y are positive integers then y^(1/x) >1



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