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Is 5^(x+2)/25<1 ?

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Is 5^(x+2)/25<1 ?  [#permalink]

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New post 25 Jun 2012, 02:07
2
11
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A
B
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D
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Re: Is 5^(x+2)/25<1 ?  [#permalink]

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New post 25 Jun 2012, 02:07
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3
SOLUTION

Is \(\frac{5^{x+2}}{25}<1\) ?

Is \(\frac{5^{x+2}}{25} <1\)? --> is \(\frac{5^{x+2}}{5^2}<1\)? --> is \(5^x<1\)? --> is \(x<0\)?

Or: is \(\frac{5^{x+2}}{25} <1\)? --> is \(5^{x+2}<5^2\)? --> is \(x+2<2\)? --> is \(x<0\)?

(1) \(5^x < 1\) --> \(x<0\). Sufficient.

(2) \(x < 0\). Directly answers the question. Sufficient.

Answer: D.
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Re: Is 5^(x+2)/25<1 ?  [#permalink]

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New post 25 Jun 2012, 07:35
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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New post 29 Jun 2012, 03:43
SOLUTION

Is \(\frac{5^{x+2}}{25}<1\) ?

Is \(\frac{5^{x+2}}{25} <1\)? --> is \(\frac{5^{x+2}}{5^2}<1\)? --> is \(5^x<1\)? --> is \(x<0\)?

Or: is \(\frac{5^{x+2}}{25} <1\)? --> is \(5^{x+2}<5^2\)? --> is \(x+2<2\)? --> is \(x<0\)?

(1) \(5^x < 1\) --> \(x<0\). Sufficient.

(2) \(x < 0\). Directly answers the question. Sufficient.

Answer: D.
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Is 5^(x+2)/25<1 ?  [#permalink]

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New post 30 Jun 2015, 01:02
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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New post 30 Jun 2015, 02:31
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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 bgpower

You 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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New post 30 Jun 2015, 02:42
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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New post 30 Jun 2015, 03:39
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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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New post 30 Jun 2015, 04:10
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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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New post 30 Jun 2015, 10:40
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.

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Re: Is 5^(x+2)/25<1 ?  [#permalink]

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New post 30 Jun 2015, 10:41
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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.

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Is 5^(x+2)/25<1 ?  [#permalink]

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New post 01 Jul 2015, 02:56
OptimusPrepJanielle

Thank 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.

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Re: Is 5^(x+2)/25<1 ?  [#permalink]

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New post 01 Jul 2015, 03:09
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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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New post 01 Jul 2015, 03:12
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bgpower wrote:
OptimusPrepJanielle

Thank 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 bgpower

When 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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New post 21 Jul 2016, 13:18
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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New post 21 Jul 2016, 13:41
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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 Witchcrafts
if 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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New post 21 Jul 2016, 14:38
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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Re: Is 5^(x+2)/25<1 ?  [#permalink]

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