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655-705 (Hard)|   Algebra|   Exponents|      
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Bunuel
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If 9^x + 9^(-x) = 62, then what is the value of 3^x + 3^(-x)? 04 Aug 2022, 02:18
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C
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65% (hard)

Question Stats:

65% (02:04) correct 35% (02:31) wrong based on 554 sessions

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If \(9^x + 9^{-x} = 62\), then what is the value of \(3^x + 3^{-x}\)?

A. 1/64
B. 1/8
C. 8
D. 62
E. 64

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C
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Bunuel
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If 9^x + 9^(-x) = 62, then what is the value of 3^x + 3^(-x)? 21 Feb 2023, 05:15
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Bunuel
If \(9^x + 9^{-x} = 62\), then what is the value of \(3^x + 3^{-x}\)?

A. 1/64
B. 1/8
C. 8
D. 62
E. 64
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Official Solution:

If \(9^x + 9^{-x} = 62\), then what is the value of \(3^x + 3^{-x}\)?

A. \(\frac{1}{64}\)
B. \(\frac{1}{8}\)
C. \(8\)
D. \(62\)
E. \(64\)


We are given that \(9^x + 9^{-x} = 62\). To find the value of \(3^x + 3^{-x}\), we can square it and simplify:

\((3^x + 3^{-x})^2 = \)

\(=(3^x)^2 + 2*3^x*3^{-x} +(3^{-x})^2=\)

\(=9^x + 2*3^x*\frac{1}{3^{x}} +9^{-x}=\)

\(=9^x + 2+ 9^{-x}\)

\(=62 + 2 = 64\)

Now, since \(3^x + 3^{-x}\) squared is 64, then \(3^x + 3^{-x}\) is 8 (notice that it cannot be -8 since \(3^x + 3^{-x}\) is positivefor all values of \(x\)).


Answer: C
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If 9^x + 9^(-x) = 62, then what is the value of 3^x + 3^(-x)? Updated on: 11 Aug 2022, 21:33
Originally posted by harshittahuja on 04 Aug 2022, 02:31.
Last edited by harshittahuja on 11 Aug 2022, 21:33, edited 1 time in total.
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3^x= a
3^-x=1/a
(a+1/a)^2=a^2+1/a^2+2*a*1/a
Implies
a^2+1/a^2+2
Taking the values
Answer is 62+2 =64
And taking the squaroot gives us 8 as answer

Posted from my mobile device
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If 9^x + 9^(-x) = 62, then what is the value of 3^x + 3^(-x)? Updated on: 06 Aug 2022, 01:49
Originally posted by DarkKnight800 on 04 Aug 2022, 02:58.
Last edited by DarkKnight800 on 06 Aug 2022, 01:49, edited 1 time in total.
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Let's write \(3^x + \frac{1}{3^x}\) as \(\sqrt{(3^x + \frac{1}{3^x})^2}\). This will make this as \(\sqrt{(9^x + \frac{1}{9^x} + 2)}\). Since we know that \(9^x + \frac{1}{9^x}\) = 62 from the question statement. Hence \(\sqrt{(9^x + \frac{1}{9^x} + 2)}\) become \(\sqrt{64}\) = 8.

Hence choice C is the correct answer.
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If 9^x + 9^(-x) = 62, then what is the value of 3^x + 3^(-x)? 04 Aug 2022, 10:12
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9^x+9^(−x) = 62
= 3^(2x) + 3^(-2x) = 62
add 2 to both sides
3^(2x) + 3^(-2x) +2 = 62+2 = 64
LHS can be written as
{3^x + 3^(-x)}^2 = 64 = 8^2
or 3^x + 3^(-x) = 8
Answer C
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If 9^x + 9^(-x) = 62, then what is the value of 3^x + 3^(-x)? 17 Sep 2023, 06:15
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Is this legal?

\(9^x + 9^{-x} = 62\\
9^x + \frac{1}{9^x}= 62 \\
\frac{1}{3^{2x}}= 62 - 3^{2x}\)
Raising both sides to 1/2 (i.e, sq. rooting both the sides)
\(3^{-x} = 62 ^ \frac{1}{2} - 3^x\\
3^x + 3^{-x} = 62 ^ \frac{1}{2}\)
This, \(3^x + 3^{-x}\) = ~\(8 \)
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If 9^x + 9^(-x) = 62, then what is the value of 3^x + 3^(-x)? 16 Oct 2024, 11:21
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Where do you get the " + 2" when you square 3^x + 3^(-x)?
DarkKnight800
Let's write \(3^x + \frac{1}{3^x}\) as \(\sqrt{(3^x + \frac{1}{3^x})^2}\). This will make this as \(\sqrt{(9^x + \frac{1}{9^x} + 2)}\). Since we know that \(9^x + \frac{1}{9^x}\) = 62 from the question statement. Hence \(\sqrt{(9^x + \frac{1}{9^x} + 2)}\) become \(\sqrt{64}\) = 8.

Hence choice C is the correct answer.
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If 9^x + 9^(-x) = 62, then what is the value of 3^x + 3^(-x)? 16 Oct 2024, 11:26
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Where do you get the " + 2" when you square 3^x + 3^(-x)?
DarkKnight800
Let's write \(3^x + \frac{1}{3^x}\) as \(\sqrt{(3^x + \frac{1}{3^x})^2}\). This will make this as \(\sqrt{(9^x + \frac{1}{9^x} + 2)}\). Since we know that \(9^x + \frac{1}{9^x}\) = 62 from the question statement. Hence \(\sqrt{(9^x + \frac{1}{9^x} + 2)}\) become \(\sqrt{64}\) = 8.

Hence choice C is the correct answer.
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That's because (a + b)^2 = a^2 + 2ab + b^2. So:


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

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

\(=9^x +2 + \frac{1}{9^x}\)
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If 9^x + 9^(-x) = 62, then what is the value of 3^x + 3^(-x)? 28 Oct 2025, 17:28
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