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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)

Sent from my LG-H818 using GMAT Club Forum mobile app
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nguyendinhtuong
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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nguyendinhtuong
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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nguyendinhtuong
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:
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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tsimo1000

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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\(\frac{1}{3^2 }\)+\(\frac{ 1}{3^3 }\)+ \(\frac{1}{3^6}\) = \(X\) * \(\frac{1}{3^5}\)

Multiply everything by 1/3^6 .... you'll get:

\(3^4 + 3^2 + 1 = 3x\)

91 = 3x

\(x = 91/3\)
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