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The value of \(3^{-2} + 3^{-4} + 3^{-6}\) is how many times the value
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25 Jan 2021, 05:31
25 Jan 2021, 05:31
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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. \((\frac{1}{3})\)
Kudos for the right answer and explanation
A. \((\frac{91}{3})\)
B. \(27\)
C. \((\frac{31}{3})\)
D. \(3\)
E. \((\frac{1}{3})\)
Kudos for the right answer and explanation
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Re: The value of \(3^{-2} + 3^{-4} + 3^{-6}\) is how many times the value
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25 Jan 2021, 06:31
25 Jan 2021, 06:31
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Carcass 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. \((\frac{1}{3})\)
Kudos for the right answer and explanation
A. \((\frac{91}{3})\)
B. \(27\)
C. \((\frac{31}{3})\)
D. \(3\)
E. \((\frac{1}{3})\)
Kudos for the right answer and explanation
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: A
Cheers,
Brent
The value of \(3^{-2} + 3^{-4} + 3^{-6}\) is how many times the value
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21 Feb 2021, 08:12
21 Feb 2021, 08:12
1
Solution:
\(3^-^2+3^-^4+3^-^6\)=\(\frac{1}{9}+\frac{1}{81}+\frac{1}{729}=\frac{90}{729}+\frac{1}{729}=\frac{91}{729} or \frac{91}{3^6}\)
Let \(3^-^5\) be x times \(\frac{91}{3^6}\)
\(\frac{1}{3}^5\) *x= \(\frac{91}{3^6}\)
Solve this to get x=\(\frac{91}{3}\)
IMO A
Hope this helps!
\(3^-^2+3^-^4+3^-^6\)=\(\frac{1}{9}+\frac{1}{81}+\frac{1}{729}=\frac{90}{729}+\frac{1}{729}=\frac{91}{729} or \frac{91}{3^6}\)
Let \(3^-^5\) be x times \(\frac{91}{3^6}\)
\(\frac{1}{3}^5\) *x= \(\frac{91}{3^6}\)
Solve this to get x=\(\frac{91}{3}\)
IMO A
Hope this helps!
