Is 5^(x+2)/25

Question

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

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

Bunuel · Accepted Answer

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.

cyberjadugar · Answer

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)

Regards,

bgpower · Answer

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 {5^1}} , but the denominator would still be always greater than the numerator?

Thank you!

Harley1980 · Answer

Hello  bgpower

You are correct that  5^\frac{-1}{10} = \frac{1}{5^(1/10)} = \frac{1}{\sqrt {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 {5^1}}  than you are wrong because in this fraction denominator  \sqrt {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 {5^1}

bgpower · Answer

Hi, Thanks for your reply!

I was actually talking about:  \frac{1}{\sqrt {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!

Harley1980 · Answer

I think you leave out last step of task (dividing on 25) and this confuse you.

You are absolutely right that  \frac{1}{\sqrt {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

bgpower · 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.

OptimusPrepJanielle · Answer

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 · Answer

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

bgpower · Answer

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!

OptimusPrepJanielle · Answer

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.

Harley1980 · Answer

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

Witchcrafts · Answer

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.

Harley1980 · Answer

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}

Witchcrafts · Answer

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

salopian · Answer

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

Hyord · Answer

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?

apurvsibalharvard · Answer

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

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

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