Solve the equation. (Enter your answers as a comma-separated list. Round your answers to four decimal places.)
2^x − 8(2^−x) + 2 = 0
2^(-x) (2^(2x) + 2*2^x - 8) = 0
2^(-x) is never zero, so that means you must have
2^(2x) + 2*2^x - 8 = 0
(2^x+4)(2^x-2)
2^x is never negative, so that means
2^x-2=0
2^x=2
x=1
To solve the equation 2^x − 8(2^−x) + 2 = 0, we can use a substitution.
Let's substitute y = 2^x. After substituting, the equation becomes:
y − 8(1/y) + 2 = 0
Multiplying through by y, we get:
y^2 - 8 + 2y = 0
Rearranging the terms, we have:
y^2 + 2y - 8 = 0
Now, we can solve this quadratic equation.
Using the quadratic formula, we have:
y = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 2, and c = -8. Plugging these values into the quadratic formula, we get:
y = (-2 ± √(2^2 - 4(1)(-8))) / (2(1))
Simplifying, we have:
y = (-2 ± √(4 + 32)) / 2
y = (-2 ± √36) / 2
y = (-2 ± 6) / 2
This gives us two possible solutions for y:
1. y = (-2 + 6) / 2 = 4 / 2 = 2
2. y = (-2 - 6) / 2 = -8 / 2 = -4
Now, we need to find the corresponding values of x.
For y = 2:
2^x = 2
Taking the logarithm base 2 of both sides, we get:
x = log2(2)
x = 1
For y = -4:
2^x = -4 (no real solution since raising 2 to any power will always give a positive value)
Therefore, the solution to the equation is x = 1.
Answer: 1
To solve the equation 2^x − 8(2^−x) + 2 = 0, we can use a substitution method. Let's make the substitution y = 2^x. Now we have the equation y − 8/y + 2 = 0.
To simplify the equation, let's multiply through by y to eliminate the denominator:
y^2 − 8 + 2y = 0
Rearranging terms, we have a quadratic equation:
y^2 + 2y − 8 = 0
To solve this quadratic equation, we can use either factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / 2a
Applying this to our equation y^2 + 2y − 8 = 0, we have:
a = 1, b = 2, c = -8
x = (-2 ± √(2^2 - 4*1*(-8))) / (2*1)
x = (-2 ± √(4 + 32)) / 2
x = (-2 ± √36) / 2
x = (-2 ± 6) / 2
Simplifying, we have:
x = (4) / 2 or x = (-8) / 2
x = 2 or x = -4
Therefore, the solutions to the equation 2^x − 8(2^−x) + 2 = 0 are x = 2 and x = -4.