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What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x  [#permalink]

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Question Stats: 20% (02:14) correct 80% (02:08) wrong based on 219 sessions

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What is the sum of all solutions to the equation $$x^{(2x^2 + 4x – 6)} = x^{(x^2 + 8x +6)}$$ ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

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Originally posted by GMATPrepNow on 02 Jan 2017, 13:45.
Last edited by GMATPrepNow on 13 Nov 2019, 16:57, edited 2 times in total.
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Re: What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x  [#permalink]

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GMATPrepNow wrote:
What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

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. For example, 1² = 1³, but we can't conclude that 2 = 3.

So, let's first see what happens when the base (x) equals 0, 1 and -1

If x = 0, then we have: 0^(2(0²) + 4(0) – 6) = 0^(0² + 8(0) + 6)
Simplify: 0^(-6) = 0^6
Evaluate: 0 = 0
So, x = 0 is one solution to the equation (yes, I know that x = 0 does not change the SUM of the solutions. I just want to show all of the possible considerations)

If x = 1, then we have: 1^(2(1²) + 4(1) – 6) = 1^(1² + 8(1) + 6)
Simplify: 1^0 = 1^15
Evaluate: 1 = 1
So, x = 1 is another solution to the equation

If x = -1, then we have: (-1)^[2(-1)² + 4(-1) – 6] = (-1)^[(-1)² + 8(-1) + 6]
Simplify: (-1)^(-8) = (-1)^(-1)
Evaluate: 1 = -1
So, x = -1 is NOT a solution to the equation

Now let's assume that x ≠ 0, x ≠ 1 and x ≠ -1 and look for other x-values that satisfy the given equation.
Given: x^(2x² + 4x – 6) = x^(x² + 8x + 6)
Since the bases are the same, we can write: 2x² + 4x – 6 = x² + 8x + 6
Rearrange to get: x² - 4x – 12 = 0
Factor to get: (x - 6)(x + 2) = 0
So, x = 6 and x = -2 are also solutions to the equation.

So, the solutions are x = 0, x = 1, x = 6, and x = -2
0 + 1 + 6 + (-2) = 5

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Re: What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x  [#permalink]

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GMATPrepNow wrote:
What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

subtract exponent 2 from exponent 1
x^2-4x-12=0
x=6,-2
6+(-2)=4
D
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Re: What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x  [#permalink]

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gracie wrote:
GMATPrepNow wrote:
What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

subtract exponent 2 from exponent 1
x^2-4x-12=0
x=6,-2
6+(-2)=4
D

That's close, but you haven't found all of the possible solutions.

Cheers,
Brent
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Re: What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x  [#permalink]

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GMATPrepNow wrote:
gracie wrote:
GMATPrepNow wrote:
What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

subtract exponent 2 from exponent 1
x^2-4x-12=0
x=6,-2
6+(-2)=4
D

That's close, but you haven't found all of the possible solutions.

Cheers,
Brent

I see it now, should have noted that 1^0=1^15.
Thanks. Good problem.
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Re: What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x  [#permalink]

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gracie wrote:
I see it now, should have noted that 1^0=1^15.
Thanks. Good problem.

Nice work.
I should also note that x = 0 is another solution (but it doesn't change the answer)

We must also check to see whether x = -1 is a possible solution.
When we plug x = -1 into the equation, we find that it is NOT a solution.

So, the solutions are x = 6, x = -2 and x = 1

Cheers,
Brent
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Re: What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x  [#permalink]

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GMATPrepNow wrote:
What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

The equation can be written as -

$$x ^{(x-6)(x+2)} = 1$$

L.H.S = R.H.S, when x = 1,6,-2
Hence, Sum = 5
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Re: What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x  [#permalink]

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Best approach would be to take log on both sides.

logx[(x-6)(x+2)] =0
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Re: What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x  [#permalink]

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srinjoy1990 wrote:
Best approach would be to take log on both sides.

logx[(x-6)(x+2)] =0

Unfortunately, you won't have a calculator (or a log table for that day) on test day.
That said, if you did have a calculator, what values would you plug in for x?

Cheers,
Brent
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Re: What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x  [#permalink]

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GMATPrepNow wrote:
srinjoy1990 wrote:
Best approach would be to take log on both sides.

logx[(x-6)(x+2)] =0

Unfortunately, you won't have a calculator (or a log table for that day) on test day.
That said, if you did have a calculator, what values would you plug in for x?

Cheers,
Brent

we need not take values thats the point, we can solve the equation by taking log.

you have something like this,

logx(x^2-4x-12) = 0.

logx =0 | x^2-4x-12=0

so, roots are, x=1 , and the solution to the equation x^2-4x-12=0.

so, 5 (1+4).
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Re: What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x  [#permalink]

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GMATPrepNow wrote:
What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?

A) -4
B) -3
C) 3
D) 4
E) 5

* Kudos for all correct solutions

x^(2x² + 4x – 6) = x^(x² + 8x +6)
2x² + 4x – 6 = x² + 8x +6
x² - 4x - 12 = (x+2)(x-6)
from here x = -2 , 6
here x = 1 will also satisfy the equation
so -2+6+1 = 5
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Re: What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x  [#permalink]

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x = 1,6,-2
Sum = 5.
Hence E.

Posted from my mobile device Re: What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x   [#permalink] 06 Nov 2019, 10:59
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