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505-555 (Easy)|   Exponents|      
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Bunuel
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If x is a positive integer, then 9^x + 3^(2x+1) = 20 Mar 2020, 02:54
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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15% (low)

Question Stats:

78% (00:58) correct 22% (01:14) wrong based on 199 sessions

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If x is a positive integer, then \(9^x + 3^{2x+1} =\)


A. \(3(3^{2x+1})\)

B. \(3^{4x−1}\)

C. \(3^{4x+1}\)

D. \(4(3^{2x})\)

E. \(12^{2x}\)
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D
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If x is a positive integer, then 9^x + 3^(2x+1) = 20 Mar 2020, 02:57
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Bunuel
If x is a positive integer, then \(9^x + 3^{2x+1} =\)


A. \(3(3^{2x+1})\)

B. \(3^{4x−1}\)

C. \(3^{4x+1}\)

D. \(4(3^{2x})\)

E. \(12^{2x}\)
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Asked: If x is a positive integer, then \(9^x + 3^{2x+1} =\)

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

IMO D
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If x is a positive integer, then 9^x + 3^(2x+1) = 20 Mar 2020, 03:10
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Bunuel
If x is a positive integer, then \(9^x + 3^{2x+1} =\)


A. \(3(3^{2x+1})\)
B. \(3^{4x−1}\)
C. \(3^{4x+1}\)
D. \(4(3^{2x})\)
E. \(12^{2x}\)
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\(9^x + 3^{2x+1}\)
= \(3^{2x} + 3*3^{2x}\)
= \(4*(3^{2x})\)

Option D
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If x is a positive integer, then 9^x + 3^(2x+1) = 28 Oct 2020, 07:48
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How do you go from 3^2x + 3^2x+1 to 3^2x (1+3).

And how do you go from 3^2x +3(3^2x) to 4(3x^2)?
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If x is a positive integer, then 9^x + 3^(2x+1) = 28 Oct 2020, 16:46
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xncm200
How do you go from 3^2x + 3^2x+1 to 3^2x (1+3).

And how do you go from 3^2x +3(3^2x) to 4(3x^2)?
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\(9^x + (3^2^x^+^1)\)

1) You can simplify the \(9^x\) to \(3^2^x\) because \(3^2\) is another way of saying 9
2) Now we have \(3^2^x + 3^2^x^+^1\)
3) Since we are adding exponents the only way to simplify is by factoring out the common expressions, which would give you this:
\(3^2^x (1 + 3)\)
The 1 is because once you factor \(3^2^x\) you only have 1 leftover. Then 3 is because of the \(^+^1\) exponent
4) If you simplify again we get:
\(4(3^2^x)\)

P.S. I'm sorry if my wording is incorrect
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If x is a positive integer, then 9^x + 3^(2x+1) = 28 Oct 2020, 19:06
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Bunuel
If x is a positive integer, then \(9^x + 3^{2x+1} =\)


A. \(3(3^{2x+1})\)

B. \(3^{4x−1}\)

C. \(3^{4x+1}\)

D. \(4(3^{2x})\)

E. \(12^{2x}\)
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Two ways

1) Take some value for x.
Let x=1, then \(9^x + 3^{2x+1} =9+3^3=36\)
Only D and E are even.
D) \(4(3^2)=4*9=36\)....YES
E) \(12^2=144\)....NO

D

2) Algebraic way
\(9^x + 3^{2x+1} =3^{2x}+3^{2x+1}=3^{2x}(1+3)=3^{2x}*4\)

D
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If x is a positive integer, then 9^x + 3^(2x+1) = 29 Oct 2020, 02:30
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\(9^x + 3^{2x+1}:\)

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

=> \((3)^{2x} [1 + 3]\)

=> \(4 (3)^{2x}\)

Answer D
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If x is a positive integer, then 9^x + 3^(2x+1) = 29 Oct 2020, 07:47
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Bunuel
If x is a positive integer, then \(9^x + 3^{2x+1} =\)


A. \(3(3^{2x+1})\)

B. \(3^{4x−1}\)

C. \(3^{4x+1}\)

D. \(4(3^{2x})\)

E. \(12^{2x}\)
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\(9^x + 3^{2x+1}\)

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

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

\(= 4*3^{2x}\), Answer must be (D)
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If x is a positive integer, then 9^x + 3^(2x+1) = 05 Nov 2020, 17:04
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Bunuel
If x is a positive integer, then \(9^x + 3^{2x+1} =\)


A. \(3(3^{2x+1})\)

B. \(3^{4x−1}\)

C. \(3^{4x+1}\)

D. \(4(3^{2x})\)

E. \(12^{2x}\)
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Solution:

Since 9^x = 3^(2x), we have:

3^(2x) + 3^(2x) * 3

3^(2x) * (1 + 3)

3^(2x) * 4

Answer: D
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If x is a positive integer, then 9^x + 3^(2x+1) = 26 Feb 2024, 06:47
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