P = the product of all x-values that satisfy the

Question

P = the product of  all  x-values that satisfy the equation  (x^2)^{(x^2 - 2x + 1)} = x^{(3x^2 + 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

GMATAcademy · Accepted Answer

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.

BrentGMATPrepNow · Answer

IMPORTANT: If b^x = b^y, then x = y,  as long as  b ≠ 0, b ≠ 1 and b ≠ -1  
For 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: ( 0 ²)^( 0 ² - 2 (0)  + 1) =  0 ^(3 (0 ²) +  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

BrentGMATPrepNow · 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!)

kanusha · Answer

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

BrentGMATPrepNow · Answer

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

UdayPratapSingh · Answer

But that give both C&E as answer

Posted from my mobile device

TariqOmar · Answer

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?

BrentGMATPrepNow · Answer

If P = 0, answer choice E doesn't work, since it says P is GREATER THAN or equal to 14

Cheers,
Brent

exc4libur · Answer

x^{2(x^2-2x+1)}=x^{(3x^2+x+8)}

base≠(0,1,-1):{2(x^2-2x+1)}={(3x^2+x+8)}…x^2+5x+6=0…(x+3)(x+2)=0 
 base≠(0,1,-1): (x+3)(x+2)=0…x=(-3,-2)

base=(0,1,-1):x^{2(x^2-2x+1)}=x^{(3x^2+x+8)}…(0,1,-1)^{anything}=(0,1,-1)^{anything} 
 base=(0,1,-1):x=(0,1,-1,anything)

P=product(-3,-2,0,1,-1,anything)=(0)(-3,-2,1,-1,anything)=0

Answer (C)

kayarat600 · Answer

How does 1 bring in the product change anything I dont get it

Krunaal · Answer

It doesn't since P is the product of all values of x, and if x = 1 is valid, 1 will be  multiplied with other values of x  you get, basis which you will have to select the correct answer where P lies.

But if x = 0 is valid, you know that P is going to be 0, and in that case looking at the answer choices, only option C. fits.

P = the product of all x-values that satisfy the

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