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

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25 Jun 2012, 03:07
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Difficulty:

5% (low)

Question Stats:

76% (00:49) correct 24% (00:49) wrong based on 940 sessions

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

(1) $$5^x < 1$$
(2) $$x < 0$$

Diagnostic Test
Question: 33
Page: 25
Difficulty: 600

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25 Jun 2012, 03:07
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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.

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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.

Regards,
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29 Jun 2012, 04:43
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.

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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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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 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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30 Jun 2015, 03:42

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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30 Jun 2015, 04:39
1
bgpower wrote:

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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30 Jun 2015, 05:10
1
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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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.

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30 Jun 2015, 11:41
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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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01 Jul 2015, 03:56
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.

Thanks!
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01 Jul 2015, 04:09
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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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01 Jul 2015, 04:12
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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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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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21 Jul 2016, 14:41
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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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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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