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What could be the value of n

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MHIKER
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What could be the value of n [#permalink] New post   Updated on: 18 Nov 2020, 05:46
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What could be the value of n?

\(\frac{1}{3}^{2-n}<\frac{1}{27}\)

A. n<-1
B. n>-1
C. n<-2
D. n<3
E. n<-3

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I'm confused with the official answer.
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Official Answer
A

Originally posted by MHIKER on 18 Nov 2020, 05:08.
Last edited by chetan2u on 18 Nov 2020, 05:46, edited 2 times in total.
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Re: What could be the value of n [#permalink] New post   18 Nov 2020, 05:39
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MHIKER wrote:
What could be the value of n?

\(\frac{1}{3}\)^2-n<\(\frac{1}{27}\)

A. n<-1
B. n>-1
C. n<-2
D. n<3
E. n<-3

I'm confused with the official answer.


\(\frac{1}{3}^{2-n}<\frac{1}{27}\)
\(\frac{1}{3}^{2-n}<{\frac{1}{3}}^3\)
It is easy to fall in the trap by equating the power as 2-n<3. But What is the BASE? -- 1/3, which is less than 1.

And what happens to numbers between 0 and 1 : They become lesser as the power increases.
So when you equate the power : 2-n>3......n<2-3 or n<-1

OR it is better to remove fraction.
\(\frac{1}{3}^{2-n}<{\frac{1}{3}}^3\)
Cross multiply as all are positive..
\(3^3<3^{2-n}\)
Now equate the power of 3
\(3<2-n....n<-1\)

A
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What could be the value of n [#permalink] New post   18 Nov 2020, 06:58
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chetan2u wrote:
MHIKER wrote:
What could be the value of n?

\(\frac{1}{3}\)^2-n<\(\frac{1}{27}\)

A. n<-1
B. n>-1
C. n<-2
D. n<3
E. n<-3

I'm confused with the official answer.


\(\frac{1}{3}^{2-n}<\frac{1}{27}\)
\(\frac{1}{3}^{2-n}<{\frac{1}{3}}^3\)
It is easy to fall in the trap by equating the power as 2-n<3. But What is the BASE? -- 1/3, which is less than 1.

And what happens to numbers between 0 and 1 : They become lesser as the power increases.
So when you equate the power: 2-n>3......n<2-3 or n<-1

OR it is better to remove fraction.
\(\frac{1}{3}^{2-n}<{\frac{1}{3}}^3\)
Cross multiply as all are positive..
\(3^3<3^{2-n}\)
Now equate the power of 3
\(3<2-n....n<-1\)

A


It is easy to fall in the trap by equating the power as 2-n<3[/b]

Yes, So I had a concept gap.
We can't equate exponents in inequality whey their base is less than 1, right? Can we do that for equality? Please explain in detail.
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Re: What could be the value of n [#permalink] New post   18 Nov 2020, 08:44
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MHIKER wrote:
What could be the value of n?

\(\frac{1}{3}^{2-n}<\frac{1}{27}\)

A. n<-1
B. n>-1
C. n<-2
D. n<3
E. n<-3

Show Spoiler
I'm confused with the official answer.

\(3^{n - 2 } < 3{^-3}\)

Or, \(n - 2 < -3\)

So, \(n < -1\), Hence (A)
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Re: What could be the value of n [#permalink] New post   19 Nov 2020, 00:09
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so mistakes aren't made with the Fractions, I find it easier to take the reciprocal of the Base and Negate the Exponent

(1/3)^ 2-n < (1/27)

(1/3)^ 2-n < (1/3)^3

---take the Reciprocal of the Base and Negate the Exponent----

(3)^ n - 2 < (3)^ -3


n - 2 < -3

n < -1

-A-
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Re: What could be the value of n [#permalink] New post   23 Nov 2020, 20:22
chetan2u wrote:
MHIKER wrote:
What could be the value of n?

\(\frac{1}{3}\)^2-n<\(\frac{1}{27}\)

A. n<-1
B. n>-1
C. n<-2
D. n<3
E. n<-3

I'm confused with the official answer.


\(\frac{1}{3}^{2-n}<\frac{1}{27}\)
\(\frac{1}{3}^{2-n}<{\frac{1}{3}}^3\)
It is easy to fall in the trap by equating the power as 2-n<3. But What is the BASE? -- 1/3, which is less than 1.

And what happens to numbers between 0 and 1 : They become lesser as the power increases.
So when you equate the power : 2-n>3......n<2-3 or n<-1

OR it is better to remove fraction.
\(\frac{1}{3}^{2-n}<{\frac{1}{3}}^3\)
Cross multiply as all are positive..
\(3^3<3^{2-n}\)
Now equate the power of 3
\(3<2-n....n<-1\)

A


Hi chetan2u,

Could you further explain how you cross multiply this inequality?

Thank you!
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Re: What could be the value of n [#permalink] New post   24 Nov 2020, 04:21
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pat93 wrote:
chetan2u wrote:
MHIKER wrote:
What could be the value of n?

\(\frac{1}{3}\)^2-n<\(\frac{1}{27}\)

A. n<-1
B. n>-1
C. n<-2
D. n<3
E. n<-3

I'm confused with the official answer.


\(\frac{1}{3}^{2-n}<\frac{1}{27}\)
\(\frac{1}{3}^{2-n}<{\frac{1}{3}}^3\)
It is easy to fall in the trap by equating the power as 2-n<3. But What is the BASE? -- 1/3, which is less than 1.

And what happens to numbers between 0 and 1 : They become lesser as the power increases.
So when you equate the power : 2-n>3......n<2-3 or n<-1

OR it is better to remove fraction.
\(\frac{1}{3}^{2-n}<{\frac{1}{3}}^3\)
Cross multiply as all are positive..
\(3^3<3^{2-n}\)
Now equate the power of 3
\(3<2-n....n<-1\)

A


Hi chetan2u,

Could you further explain how you cross multiply this inequality?

Thank you!


\(\frac{1}{3}^{2-n}<{\frac{1}{3}}^3\)
\(\frac{1^{2-n}}{3^{2-n}}<{\frac{1^3}{3^3}}\)

\(\frac{1}{3^{2-n}}<{\frac{1}{3^3}}\)

Cross multiply
\(1^{2-n}*3^3<1^3*3^{2-n}..........3^3<3^{2-n}.\)
Now equate the powers to get 3<2-n.....n<-1
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Re: What could be the value of n [#permalink] New post   15 Aug 2023, 21:11
MHIKER wrote:
What could be the value of n?

\(\frac{1}{3}^{2-n}<\frac{1}{27}\)

A. n<-1
B. n>-1
C. n<-2
D. n<3
E. n<-3

Show Spoiler
I'm confused with the official answer.



Problem:

Any number less than -3 will also be less than -1

So we have two correct answers A and E since the question asks which COULD be the value of n.

n = -4 works

Posted from my mobile device
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Re: What could be the value of n [#permalink] New post   31 Aug 2023, 04:03
Not visible from the mobile and iPad, this should be written as (1/3)^ 2-n < (1/27)

Posted from my mobile device
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Re: What could be the value of n [#permalink] New post   31 Aug 2023, 06:22
MHIKER wrote:
What could be the value of n?

\(\frac{1}{3}^{2-n}<\frac{1}{27}\)

A. n<-1
B. n>-1
C. n<-2
D. n<3
E. n<-3

Show Spoiler
I'm confused with the official answer.


When 1/m > 1/n
It means m < n

Applying same logic
3^2-n > 3^3

Since base is positive,
Powers must be have same inequality

2-n > 3
n <-1

Posted from my mobile device
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Re: What could be the value of n [#permalink] New post   31 Aug 2023, 10:32
(1/3)^2-n < (1/27)

Cross multiplying,
=> 27 < 3^(2-n)
=> 3^3 < 3^(2-n)

Comparing powers,

3 < 2 - n
=> n < -1

Hence A
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Re: What could be the value of n [#permalink]
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