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P = the product of all xvalues that satisfy the
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Updated on: 09 Jan 2017, 13:32
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P = the product of all xvalues that satisfy the equation (x²)^(x²  2x + 1) = x^(3x² + x + 8) Which of the following is true? A) P < 14 B) 14 ≤ P < 4 C) 4 ≤ P < 4 D) 4 ≤ P < 14 E) 14 ≤ P *kudos for all correct solutions
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Originally posted by GMATPrepNow on 09 Jan 2017, 13:01.
Last edited by GMATPrepNow on 09 Jan 2017, 13:32, edited 1 time in total.



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Re: P = the product of all xvalues that satisfy the
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09 Jan 2017, 13:27
With complicated equations, especially those that include ugly exponents, it can be helpful to consider easy cases first, often 0 and 1.
Here, 1 satisfies the equation (zero doesn't since we end up with zero raised to a negative, which is the same as dividing by zero). So our answer needs to include 1.
C is the only option that includes 1, so that should be our answer.



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Re: P = the product of all xvalues that satisfy the
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09 Jan 2017, 13:34
GMATAcademy wrote: With complicated equations, especially those that include ugly exponents, it can be helpful to consider easy cases first, often 0 and 1.
Here, 1 satisfies the equation (zero doesn't since we end up with zero raised to a negative, which is the same as dividing by zero). So our answer needs to include 1.
C is the only option that includes 1, so that should be our answer. Argh! You're totally right about zero raised to a negative. I have edited the question to reflect the intent of my question (the correct answer is still C though!)
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Re: P = the product of all xvalues that satisfy the
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10 Jan 2017, 11:33
GMATPrepNow wrote: P = the product of all xvalues that satisfy the equation (x²)^(x²  2x + 1) = x^(3x² + x + 8) Which of the following is true?
A) P < 14 B) 14 ≤ P < 4 C) 4 ≤ P < 4 D) 4 ≤ P < 14 E) 14 ≤ P
*kudos for all correct solutions IMPORTANT: If b^x = b^y, then x = y, as long as b ≠ 0, b ≠ 1 and b ≠ 1For example, if we have 1^x = 1^y, we cannot conclude that x = y, since 1^x equals 1^y FOR ALL values of x and y. So, although 1² = 1³, we can't then conclude that 2 = 3. So, let's first see what happens when the base (x) equals 0. If x = 0, then we get: ( 0²)^( 0²  2 (0) + 1) = 0^(3 (0²) + 0 + 8) Simplify: 0^1 = 0^8 Evaluate: 0 = 0 Perfect! We know that x = 0 is one solution to the equation. This means the PRODUCT of all of the solutions will be ZERO, regardless of the other solutions. In other words, P = 0 So, the correct answer is C Cheers, Brent
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Re: P = the product of all xvalues that satisfy the
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12 Jan 2017, 10:10
I simplified equation, results two x values \((x^{2}) ^{x^{2}  2x + 1}\) = \(x^{3x^{2} + x + 8}\) On simplifying \(0 = x^{2} + 5x + 6\) x = 2 ; x = 3 ?? Could anyone pls explain.Thanks
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Re: P = the product of all xvalues that satisfy the
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12 Jan 2017, 12:25
kanusha wrote: I simplified equation, results two x values \((x^{2}) ^{x^{2}  2x + 1}\) = \(x^{3x^{2} + x + 8}\) On simplifying \(0 = x^{2} + 5x + 6\) x = 2 ; x = 3 ?? Could anyone pls explain.Thanks That's great work, but you need to keep going. We also need to test whether the equation also holds true when the base (x) equals 0, 1 and 1 See my earlier post for this. Once we've checked those values as well, we see that all possible solutions are x = 2, 3, 0, 1 and 1 So, P = (2)(3)(0)(1)(1) = 0
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Re: P = the product of all xvalues that satisfy the
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24 Dec 2018, 21:10
GMATPrepNow wrote: GMATPrepNow wrote: P = the product of all xvalues that satisfy the equation (x²)^(x²  2x + 1) = x^(3x² + x + 8) Which of the following is true?
A) P < 14 B) 14 ≤ P < 4 C) 4 ≤ P < 4 D) 4 ≤ P < 14 E) 14 ≤ P
*kudos for all correct solutions IMPORTANT: If b^x = b^y, then x = y, as long as b ≠ 0, b ≠ 1 and b ≠ 1For example, if we have 1^x = 1^y, we cannot conclude that x = y, since 1^x equals 1^y FOR ALL values of x and y. So, although 1² = 1³, we can't then conclude that 2 = 3. So, let's first see what happens when the base (x) equals 0. If x = 0, then we get: ( 0²)^( 0²  2 (0) + 1) = 0^(3 (0²) + 0 + 8) Simplify: 0^1 = 0^8 Evaluate: 0 = 0 Perfect! We know that x = 0 is one solution to the equation. This means the PRODUCT of all of the solutions will be ZERO, regardless of the other solutions. In other words, P = 0 So, the correct answer is C Cheers, Brent But that give both C&E as answer Posted from my mobile device



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Re: P = the product of all xvalues that satisfy the
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25 Dec 2018, 03:26
Hi,
If I simplify the equations above, I will arrive at: X^2+5x+6 = 0
The product of the solutions can be determined by the rule that the product equals = c/a when the equation is in the form ax^2+bx+c=0. Using this will lead us to the product being 6.
Can anyone from the experts tell what went wrong here please?



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Re: P = the product of all xvalues that satisfy the
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30 Dec 2018, 10:35
UdayPratapSingh wrote: GMATPrepNow wrote: GMATPrepNow wrote: P = the product of all xvalues that satisfy the equation (x²)^(x²  2x + 1) = x^(3x² + x + 8) Which of the following is true?
A) P < 14 B) 14 ≤ P < 4 C) 4 ≤ P < 4 D) 4 ≤ P < 14 E) 14 ≤ P
*kudos for all correct solutions IMPORTANT: If b^x = b^y, then x = y, as long as b ≠ 0, b ≠ 1 and b ≠ 1For example, if we have 1^x = 1^y, we cannot conclude that x = y, since 1^x equals 1^y FOR ALL values of x and y. So, although 1² = 1³, we can't then conclude that 2 = 3. So, let's first see what happens when the base (x) equals 0. If x = 0, then we get: ( 0²)^( 0²  2 (0) + 1) = 0^(3 (0²) + 0 + 8) Simplify: 0^1 = 0^8 Evaluate: 0 = 0 Perfect! We know that x = 0 is one solution to the equation. This means the PRODUCT of all of the solutions will be ZERO, regardless of the other solutions. In other words, P = 0 So, the correct answer is C Cheers, Brent But that give both C&E as answer Posted from my mobile deviceIf P = 0, answer choice E doesn't work, since it says P is GREATER THAN or equal to 14 Cheers, Brent
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Re: P = the product of all xvalues that satisfy the
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