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 ≠ -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. 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 =

-20 +

1 +

6 + (

-2) = 5

Answer: E

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