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25 Jul 2024, 01:41
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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08 Dec 2024, 02:35
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OE
In this question, you are asked to compare \( 2^{x}\) with \(3^{x+1}\), given that x is a negative integer. One way to approach this problem is to plug a value of x in both expressions and compare the results.
You are given that x is a negative integer, so the greatest integer you can plug in for x is -1.
For x = -1, it follows that \( 2^{x} = 2^{-1}=\frac{1}{2}\) and \(3^{x+1} = 3^0 = 1\).
In this case, \(2^{x}\) is less than \(3^{x+1}\). However, to conclude that Quantity B is greater, it is not sufficient for \(2^{x}\) to be less than \(3^{x+1}\) for one particular value of x; the relationship would need to be true for all negative integer values of x. To analyze this relationship further, plug in another value of x, for example, -2.
For x = -2, it follows that \(2^x = 2^{-2} = \frac{1}{2^2}=\frac{1}{4}\) and \(3^{x+1} = 3^{-2+1} = 3^{-1} = \frac{1}{3}\).
Again, \( 2^{x}\) is less than \(3^{x+1}\), but note that these values are closer together than the previous values of \( 2^{x}\) and \(3^{x+1}\). It appears that the relationship between the quantities may differ for smaller values of x, so now try plugging in -3 for x.
For x = -3, it follows that \(2^x=2^{-3} = \frac{1}{2^3} = \frac{1}{8}\) and \(3^{x+1}= 3^{-3+1} = 3^-2 = \frac{2}{3^2}=\frac{1}{9}\). In this case, \( 2^{x}\) is greater than \(3^{x+1}\).
Since is less than for x = -1 and is greater than for x = -3, the relationship between these two quantities cannot be determined from the information given.
The correct answer is Choice D.
In this question, you are asked to compare \( 2^{x}\) with \(3^{x+1}\), given that x is a negative integer. One way to approach this problem is to plug a value of x in both expressions and compare the results.
You are given that x is a negative integer, so the greatest integer you can plug in for x is -1.
For x = -1, it follows that \( 2^{x} = 2^{-1}=\frac{1}{2}\) and \(3^{x+1} = 3^0 = 1\).
In this case, \(2^{x}\) is less than \(3^{x+1}\). However, to conclude that Quantity B is greater, it is not sufficient for \(2^{x}\) to be less than \(3^{x+1}\) for one particular value of x; the relationship would need to be true for all negative integer values of x. To analyze this relationship further, plug in another value of x, for example, -2.
For x = -2, it follows that \(2^x = 2^{-2} = \frac{1}{2^2}=\frac{1}{4}\) and \(3^{x+1} = 3^{-2+1} = 3^{-1} = \frac{1}{3}\).
Again, \( 2^{x}\) is less than \(3^{x+1}\), but note that these values are closer together than the previous values of \( 2^{x}\) and \(3^{x+1}\). It appears that the relationship between the quantities may differ for smaller values of x, so now try plugging in -3 for x.
For x = -3, it follows that \(2^x=2^{-3} = \frac{1}{2^3} = \frac{1}{8}\) and \(3^{x+1}= 3^{-3+1} = 3^-2 = \frac{2}{3^2}=\frac{1}{9}\). In this case, \( 2^{x}\) is greater than \(3^{x+1}\).
Since is less than for x = -1 and is greater than for x = -3, the relationship between these two quantities cannot be determined from the information given.
The correct answer is Choice D.
