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The value of 3^(-2)+3^(-4)+3^(-6) is how many times the value of

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The value of 3^(-2)+3^(-4)+3^(-6) is how many times the value of  [#permalink]

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New post 11 Apr 2017, 07:06
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The value of \(3^{-2} + 3^{-4} + 3^{-6}\) is how many times the value of \(3^{-5}\)?

A. \(\frac{91}{3}\)

B. 27

C. \(\frac{31}{3}\)

D. 3

E. \(\frac{1}{3}\)

Source: GMAT Free
Spoiler: OA

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Re: The value of 3^(-2)+3^(-4)+3^(-6) is how many times the value of  [#permalink]

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New post 11 Apr 2017, 07:08
nguyendinhtuong wrote:
The value of \(3^{-2} + 3^{-4} + 3^{-6}\) is how many times the value of \(3^{-5}\)?

A. \(\frac{91}{3}\)

B. 27

C. \(\frac{31}{3}\)

D. 3

E. \(\frac{1}{3}\)

Source: GMAT Free


Similar question: https://gmatclub.com/forum/the-value-of ... 30682.html
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Re: The value of 3^(-2)+3^(-4)+3^(-6) is how many times the value of  [#permalink]

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New post 11 Apr 2017, 07:50
3^-2 = 1/3^2
Hence using this logic, rewriting the above equation, we get
1/3^2 + 1/3^4 + 1/3^6
= (3^4 + 3^2 + 1)/3^6
= 91/3^6
This is 91/3 times 1/3^5(Option A)

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Re: The value of 3^(-2)+3^(-4)+3^(-6) is how many times the value of  [#permalink]

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New post 11 Apr 2017, 08:29
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nguyendinhtuong wrote:
The value of \(3^{-2} + 3^{-4} + 3^{-6}\) is how many times the value of \(3^{-5}\)?

A. \(\frac{91}{3}\)

B. 27

C. \(\frac{31}{3}\)

D. 3

E. \(\frac{1}{3}\)

Source: GMAT Free


You can divide (3^-2 + 3^-4 + 3^-6) by 3^-5

We get: 3^-2+5 + 3^1 + 3^-1 = 27 + 3 + 1/3 = 30 + 1/3 = 91/3 Hence A
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Re: The value of 3^(-2)+3^(-4)+3^(-6) is how many times the value of  [#permalink]

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New post 11 Apr 2017, 12:19
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nguyendinhtuong wrote:
The value of \(3^{-2} + 3^{-4} + 3^{-6}\) is how many times the value of \(3^{-5}\)?

A. \(\frac{91}{3}\)

B. 27

C. \(\frac{31}{3}\)

D. 3

E. \(\frac{1}{3}\)

Source: GMAT Free


\(3^{-2} + 3^{-4} + 3^{-6}\)

= \(3^{-2} ( 1 + 3^{-2} + 3^{-4}\) )

= \(\frac{1}{9} ( 1 + \frac{1}{9} + \frac{1}{81} )\)

= \(\frac{1}{9} ( \frac{81 + 9 + 1}{81} )\)

= \(\frac{91}{729}\)


Quote:
is how many times the value of \(3^{-5}\)?


So, \(\frac{91}{729}*\frac{243}{1}\) = \(\frac{91}{3}\)

Hence, answer must be (A) \(\frac{91}{3}\)
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Re: The value of 3^(-2)+3^(-4)+3^(-6) is how many times the value of  [#permalink]

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New post 11 Apr 2017, 12:29
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nguyendinhtuong wrote:
The value of \(3^{-2} + 3^{-4} + 3^{-6}\) is how many times the value of \(3^{-5}\)?

A. \(\frac{91}{3}\)

B. 27

C. \(\frac{31}{3}\)

D. 3

E. \(\frac{1}{3}\)

Source: GMAT Free


Let's factor out \(3^{-5}\)

We get: \(3^{-2} + 3^{-4} + 3^{-6} = 3^{-5}(3^3 + 3^1 + 3^{-1})\)

\(= 3^{-5}(27 + 3 + \frac{1}{3})\)

\(= 3^{-5}(30 + \frac{1}{3})\)

\(= 3^{-5}(\frac{90}{3} + \frac{1}{3})\)

\(= 3^{-5}(\frac{91}{3})\)

Answer:
Spoiler: ::
A


Cheers,
Brent
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Re: The value of 3^(-2)+3^(-4)+3^(-6) is how many times the value of  [#permalink]

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New post 23 Jan 2019, 21:31
GMATPrepNow wrote:
nguyendinhtuong wrote:
The value of \(3^{-2} + 3^{-4} + 3^{-6}\) is how many times the value of \(3^{-5}\)?

A. \(\frac{91}{3}\)

B. 27

C. \(\frac{31}{3}\)

D. 3

E. \(\frac{1}{3}\)

Source: GMAT Free


Let's factor out \(3^{-5}\)

We get: \(3^{-2} + 3^{-4} + 3^{-6} = 3^{-5}(3^3 + 3^1 + 3^{-1})\)

\(= 3^{-5}(27 + 3 + \frac{1}{3})\)

\(= 3^{-5}(30 + \frac{1}{3})\)

\(= 3^{-5}(\frac{90}{3} + \frac{1}{3})\)

\(= 3^{-5}(\frac{91}{3})\)

Answer:
Spoiler: ::
A


Cheers,
Brent


can you please explain how did we get positive exponents in 3^{-5}(3^3 + 3^1 + 3^{-1})[/m] ?
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Re: The value of 3^(-2)+3^(-4)+3^(-6) is how many times the value of  [#permalink]

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New post 24 Jan 2019, 07:00
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Top Contributor
tsimo1000 wrote:

can you please explain how did we get positive exponents in 3^{-5}(3^3 + 3^1 + 3^{-1})[/m] ?


When we multiply two powers with the same base, we ADD the exponents.
For example: (k^3)(k^5) = k^8
And (7^10)(7^2) = 7^12

The same applies with NEGATIVE exponents. For example:
[x^(-3)][x^7] = x^4

The same applies to the original question.

That is, if we take \(3^{-5}(3^3 + 3^1 + 3^{-1})\) and we EXPAND it, we get \(3^{-2} + 3^{-4} + 3^{-6}\)

Does that help?

Cheers,
Brent
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Re: The value of 3^(-2)+3^(-4)+3^(-6) is how many times the value of   [#permalink] 24 Jan 2019, 07:00
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