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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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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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Thank you very much for reading this post till the end! Kudos?

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?

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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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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Thank you very much for reading this post till the end! Kudos?

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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\(\frac{5^(x+2)}{25}<1\) \(\frac{(5^x)(5^2)}{5^2}<1\) => 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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Thank you very much for reading this post till the end! Kudos?

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.

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.

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!
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Thank you very much for reading this post till the end! Kudos?

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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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.

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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