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

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.

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 \(\sqrt[10]{5^1}\)

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

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.

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

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

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!

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.

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 \(5^{x+2} < 5^2\)

For\frac{\(5^(x+2)[}{m]/25} < 1, the numerator will have to be less than 25. \\
\\
[m]5^(x+2)\) < 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 \(5^x = \sqrt{5}\)

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

Hi - how did you go from 5^x<1--> x<0?

Why would it not be x<1? Because is 1 also not just 1^1, thus 5^x<1^1 --> x<1?

Many thanks

Hi, for (1). 5^x < 1

Can we modify this to be 5^x < 5^0, then solve using the equal exponents theorem to say the equation is now x < 0?

5^(1/10) = 1.17461894309. Therefore yes, the denominator would be greater than the numerator if even it is a negative fractional value.