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If x not equal to 0, then (1/x)^n - (1/x)^(n+1)

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Bunuel
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If x not equal to 0, then (1/x)^n - (1/x)^(n+1) [#permalink] New post   21 Oct 2021, 03:45
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76% (01:26) correct 24% (01:28) wrong based on 104 sessions
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If x not equal to 0, then ( \(\frac{1}{x})^n- (\frac{1}{x})^{n+1}=\)

(A) \(\frac{-1}{x}\)

(B) \(\frac{1}{x}\)

(C) \(\frac{1}{x^n}\)

(D) \(\frac{x-1}{x^n}\)

(E) \(\frac{x-1}{x^{ n+1}}\)
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E
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BrentGMATPrepNow
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Re: If x not equal to 0, then (1/x)^n - (1/x)^(n+1) [#permalink] New post   21 Oct 2021, 08:34
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Bunuel wrote:
If x not equal to 0, then ( \(\frac{1}{x})^n- (\frac{1}{x})^{n+1}=\)

(A) \(\frac{-1}{x}\)

(B) \(\frac{1}{x}\)

(C) \(\frac{1}{x^n}\)

(D) \(\frac{x-1}{x^n}\)

(E) \(\frac{x-1}{x^{ n+1}}\)


Factor to get: \((\frac{1}{x})^n- (\frac{1}{x})^{n+1}=(\frac{1}{x})^n[1 - \frac{1}{x}]\)

Rewrite \(1\) as \(\frac{x}{x}\) to get: \((\frac{1}{x})^n- (\frac{1}{x})^{n+1}=(\frac{1}{x})^n[\frac{x}{x} - \frac{1}{x}]\)

Combine fractions to get: \((\frac{1}{x})^n- (\frac{1}{x})^{n+1}=(\frac{1}{x})^n[\frac{x-1}{x}]\)

Rewrite as follows: : \((\frac{1}{x})^n- (\frac{1}{x})^{n+1}=(\frac{1^n}{x^n})[\frac{x-1}{x}]\)

Expand: \((\frac{1}{x})^n- (\frac{1}{x})^{n+1}=\frac{(1^n)(x-1)}{(x^n)(x)}\)

Simplify: \((\frac{1}{x})^n- (\frac{1}{x})^{n+1}=\frac{x-1}{x^{ n+1}}\)

Answer: E
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chetan2u
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Re: If x not equal to 0, then (1/x)^n - (1/x)^(n+1) [#permalink] New post   21 Oct 2021, 08:50
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Bunuel wrote:
If x not equal to 0, then ( \(\frac{1}{x})^n- (\frac{1}{x})^{n+1}=\)

(A) \(\frac{-1}{x}\)

(B) \(\frac{1}{x}\)

(C) \(\frac{1}{x^n}\)

(D) \(\frac{x-1}{x^n}\)

(E) \(\frac{x-1}{x^{ n+1}}\)



I would do as BrentGMATPrepNow or I would pick some smart number for n.

Say n=-1, ( \(\frac{1}{x})^{(-1)}- (\frac{1}{x})^{-1+1}=x^1-(\frac{1}{x})^0=x-1\)

Check the choices.


(A) \(\frac{-1}{x}\neq x-1\)

(B) \(\frac{1}{x}\neq x-1\)

(C) \(\frac{1}{x^n}=\frac{1}{x{-1}}=x\neq -1\)

(D) \(\frac{x-1}{x^n}=\frac{x-1}{x^{-1}}=x(x-1)\neq x-1\)

(E) \(\frac{x-1}{x^{ n+1}} =\frac{x-1}{x^{-1+1}}=(x-1)\)…..Correct


E
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Re: If x not equal to 0, then (1/x)^n - (1/x)^(n+1) [#permalink] New post   21 Oct 2021, 08:51
Let's use some numbers put x=2 and n=2

By doing some basic calculations we see that the result of the equation given is 1/8 ,by checking all the other options we realize that only answer choice E gives us 1/8
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Re: If x not equal to 0, then (1/x)^n - (1/x)^(n+1) [#permalink] New post   21 Oct 2021, 10:09
BrentGMATPrepNow wrote:
Bunuel wrote:
If x not equal to 0, then ( \(\frac{1}{x})^n- (\frac{1}{x})^{n+1}=\)

(A) \(\frac{-1}{x}\)

(B) \(\frac{1}{x}\)

(C) \(\frac{1}{x^n}\)

(D) \(\frac{x-1}{x^n}\)

(E) \(\frac{x-1}{x^{ n+1}}\)


Factor to get: \((\frac{1}{x})^n- (\frac{1}{x})^{n+1}=(\frac{1}{x})^n[1 - \frac{1}{x}]\)

Rewrite \(1\) as \(\frac{x}{x}\) to get: \((\frac{1}{x})^n- (\frac{1}{x})^{n+1}=(\frac{1}{x})^n[\frac{x}{x} - \frac{1}{x}]\)

Combine fractions to get: \((\frac{1}{x})^n- (\frac{1}{x})^{n+1}=(\frac{1}{x})^n[\frac{x-1}{x}]\)

Rewrite as follows: : \((\frac{1}{x})^n- (\frac{1}{x})^{n+1}=(\frac{1^n}{x^n})[\frac{x-1}{x}]\)

Expand: \((\frac{1}{x})^n- (\frac{1}{x})^{n+1}=\frac{(1^n)(x-1)}{(x^n)(x)}\)

Simplify: \((\frac{1}{x})^n- (\frac{1}{x})^{n+1}=\frac{x-1}{x^{ n+1}}\)

Answer: E



Would it not be
(1/x)^n - (1/x)^n* (1/x)^1 = -(1/x)^1
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BrentGMATPrepNow
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Re: If x not equal to 0, then (1/x)^n - (1/x)^(n+1) [#permalink] New post   21 Oct 2021, 10:30
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JyotiChauhan wrote:
Would it not be
(1/x)^n - (1/x)^n* (1/x)^1 = -(1/x)^1


No that's not correct.
(1/x)^n - (1/x)^n* (1/x)^1 doesn't equal -(1/x)^1

You can test it by plugging in values of x and n
For example, if x = 1 and n = 1, then (1/x)^n - (1/x)^n* (1/x)^1 = (1/1)^1 - (1/1)^1 * (1/1)^1 = 1 - 1 = 0

However, if x = 1 and n = 1, then -(1/x)^1 = -(1/1)^1 = -1

So, as you can see (1/x)^n - (1/x)^n* (1/x)^1 doesn't equal -(1/x)^1
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Re: If x not equal to 0, then (1/x)^n - (1/x)^(n+1) [#permalink]
21 Oct 2021, 10:30
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