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

Answer: E
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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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I see it now, should have noted that 1^0=1^15.
Add 1 to 4 and the answer is 5.
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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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

Best approach would be to take log on both sides.

logx[(x-6)(x+2)] =0
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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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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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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
E is the answer

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

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