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605-655 (Medium)|   Algebra|   Exponents|      
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Ekland
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3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6 is how many times 3^(-14) ? Updated on: 15 Feb 2016, 07:22
Originally posted by Ekland on 15 Feb 2016, 07:02.
Last edited by Bunuel on 15 Feb 2016, 07:22, edited 1 time in total.
Renamed the topic and edited the question.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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45% (medium)

Question Stats:

75% (02:46) correct 25% (02:44) wrong based on 684 sessions

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\((\frac{3^{-11}}{2}) + (\frac{3^{-12}}{4}) + (\frac{3^{-13}}{6})\) is how many times \(3^{-14}\) ?

A. 1/3
B. 21
C. 65/4
D. 24
E. 251/6
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Official Answer
C
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Bunuel
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3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6 is how many times 3^(-14) ? 15 Feb 2016, 07:39
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Nez
\((\frac{3^{-11}}{2}) + (\frac{3^{-12}}{4}) + (\frac{3^{-13}}{6})\) is how many times \(3^{-14}\) ?

A. 1/3
B. 21
C. 65/4
D. 24
E. 251/6
Show more

\(\frac{(\frac{3^{-11}}{2}) + (\frac{3^{-12}}{4}) + (\frac{3^{-13}}{6})}{3^{-14}}=\)

\(=(\frac{3^{3}}{2}) + (\frac{3^{2}}{4}) + (\frac{3}{6})=\)

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

\(=(\frac{2*3^{3}}{4}) + (\frac{3^{2}}{4}) + (\frac{2}{4})=\frac{65}{4}\)

Answer: C.
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3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6 is how many times 3^(-14) ? 15 Feb 2016, 07:35
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Nez
\((\frac{3^{-11}}{2}) + (\frac{3^{-12}}{4}) + (\frac{3^{-13}}{6})\) is how many times \(3^{-14}\) ?

A. 1/3
B. 21
C. 65/4
D. 24
E. 251/6
Show more


You are asked for the value of x such that \(x*3^{-14} = (\frac{3^{-11}}{2}) + (\frac{3^{-12}}{4}) + (\frac{3^{-13}}{6})\)

In order to find x, lets focus on\((\frac{3^{-11}}{2}) + (\frac{3^{-12}}{4}) + (\frac{3^{-13}}{6})= (\frac{2*3^{-10}+3^{-11}+2*3^{-13}}{12}) = \frac{3^{-14}*195}{12} = \frac{3^{-14}*65}{4}\)

Thus, \(x = \frac {65}{4}\)

C is thus the correct answer.

Hope this helps.
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shukiPortal12
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3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6 is how many times 3^(-14) ? Updated on: 21 Dec 2016, 10:55
Originally posted by shukiPortal12 on 21 Dec 2016, 10:22.
Last edited by shukiPortal12 on 21 Dec 2016, 10:55, edited 1 time in total.
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Can you please explain how did you get from (\frac{3^{-11}}{2}) + (\frac{3^{-12}}{4}) + (\frac{3^{-13}}{6}) to (\frac{2*3^{-10}+3^{-11}+2*3^{-13}}{12})
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3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6 is how many times 3^(-14) ? 21 Dec 2016, 10:33
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shukiPortal12
Can you please how did you get from (\frac{3^{-11}}{2}) + (\frac{3^{-12}}{4}) + (\frac{3^{-13}}{6}) to (\frac{2*3^{-10}+3^{-11}+2*3^{-13}}{12})
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There are TWO steps between those two equations, which one is unclear?

\(\frac{(\frac{3^{-11}}{2}) + (\frac{3^{-12}}{4}) + (\frac{3^{-13}}{6})}{3^{-14}}=\)

\(=(\frac{3^{3}}{2}) + (\frac{3^{2}}{4}) + (\frac{3}{6})=\)

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

\(=(\frac{2*3^{3}}{4}) + (\frac{3^{2}}{4}) + (\frac{2}{4})=\frac{65}{4}\)


Also, please format properly so that it's clear what you mean: rules-for-posting-please-read-this-before-posting-133935.html#p1096628
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3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6 is how many times 3^(-14) ? 21 Dec 2016, 11:00
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from the 1st, (3−112)+(3−124)+(3−136)3−14=(3−112)+(3−124)+(3−136)3−14
to
(332)+(324)+(36)

I never get to understand whether there's a way to add distinct powers with distinct bases (such as 3^7 + 2^11)

Sorry about my English. It's not so good...
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3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6 is how many times 3^(-14) ? 21 Dec 2016, 15:30
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Dumb question but how do we go from (3^-11)/2 to (3^3)/2? That's not totally clear to me.
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3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6 is how many times 3^(-14) ? 22 Dec 2016, 02:05
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shukiPortal12
from the 1st, (3−112)+(3−124)+(3−136)3−14=(3−112)+(3−124)+(3−136)3−14
to
(332)+(324)+(36)

I never get to understand whether there's a way to add distinct powers with distinct bases (such as 3^7 + 2^11)

Sorry about my English. It's not so good...
Show more

Added few more steps:

\(\frac{(\frac{3^{-11}}{2}) + (\frac{3^{-12}}{4}) + (\frac{3^{-13}}{6})}{3^{-14}}=\)

\(=\frac{3^{-11}}{2*3^{-14}} + \frac{3^{-12}}{4*3^{-14}} + \frac{3^{-13}}{6*3^{-14}}=\)

\(=\frac{3^{-11-(-14)}}{2} + \frac{3^{-12-(-14)}}{4} + \frac{3^{-13-(-14)}}{6}=\)

\(=(\frac{3^{3}}{2}) + (\frac{3^{2}}{4}) + (\frac{3}{6})=\)

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

\(=(\frac{2*3^{3}}{4}) + (\frac{3^{2}}{4}) + (\frac{2}{4})=\frac{65}{4}\).

Hope it's clear.
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3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6 is how many times 3^(-14) ? 22 Dec 2016, 07:28
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Very clear. Thank you very much.
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3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6 is how many times 3^(-14) ? 12 Apr 2017, 19:52
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In order to solve this question we need to set up an algebraic expression:

3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6= x (3^-14)

Trick* if remember that any fraction divided by a number is that fraction multiplied by the reciprocal of that number ( e.x (1/2)/3=(1/2) * (1/3)) then we can rewrite the expression as so:

3^(-11)/2(3^-14) + 3^(-12)/4(3^-14) + 3^(-13)/6(3^-14)= x
3^3/2 + 3^2/2^2 + 3/6=
27/2 + 9/4 + 3/6 = 65/4
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3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6 is how many times 3^(-14) ? 22 Nov 2020, 12:01
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Ekland
\((\frac{3^{-11}}{2}) + (\frac{3^{-12}}{4}) + (\frac{3^{-13}}{6})\) is how many times \(3^{-14}\) ?

A. 1/3
B. 21
C. 65/4
D. 24
E. 251/6
Show more

= \(\frac{(\frac{3^{-11}}{2}) + (\frac{3^{-12}}{4}) + (\frac{3^{-13}}{6})}{3^{-14}}=\)

= \(\frac{3^{11-(-14)}}{2} + \frac{3^{-12-(-14)}}{4} + \frac{3^{-12-(-14)}}{6}\)

= \(\frac{3^3}{2} + \frac{3^2}{4} + \frac{3}{6}\)

= \(\frac{65}{4}\)
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3^(-11)/2 + 3^(-12)/4 + 3^(-13)/6 is how many times 3^(-14) ? 29 Nov 2025, 11:07
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