What is the solution set of 4^(x^(2)-4x)=2^(-6)
4^(x^2 - 4x) = 2^-6
(2^2)^(x^2-4x) = 2^-6
2^(2x^2 - 8x) = 2^-6
then
2x^2 - 8x = -6
x^2 - 4x + 3 = 0
(x-1)(x-3) = 0
x = 1 or x = 3
Thanks!
To find the solution set of the equation 4^(x^2 - 4x) = 2^(-6), we can first rewrite both sides of the equation using the same base. Since 4 can be expressed as 2^2, the equation becomes:
(2^2)^(x^2 - 4x) = 2^(-6)
Applying the exponent rule that says (a^b)^c = a^(b*c), we can further simplify the equation:
2^(2*(x^2 - 4x)) = 2^(-6)
Now, since both sides have the same base (2), we can equate the exponents:
2*(x^2 - 4x) = -6
Next, solve this equation by distributing the 2 on the left side:
2x^2 - 8x = -6
Then, bring all terms to one side of the equation to obtain a quadratic equation:
2x^2 - 8x + 6 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 2, b = -8, and c = 6. Substituting these values into the formula:
x = (-(-8) ± √((-8)^2 - 4*2*6)) / (2*2)
x = (8 ± √(64 - 48)) / 4
x = (8 ± √16) / 4
Now, simplify the square root:
x = (8 ± 4) / 4
This results in two possible solutions:
x = (8 + 4) / 4 = 12 / 4 = 3
x = (8 - 4) / 4 = 4 / 4 = 1
Therefore, the solution set of the equation 4^(x^2 - 4x) = 2^(-6) is {1, 3}.