how do you solve 2(2^2x)-5(2^x)+2=0 for x
Let 2^x = y
then your equation becomes
2y^2 - 5y + 2 = 0
(2y-1)(y-2) = 0
y = 1/2 or y = 2
then 2^x = 1/2 --- x = -1
or
2^x = 2 --- x = 1
To solve the equation 2(2^(2x))-5(2^x)+2=0 for x, we can use a substitution to make the equation easier to handle.
Let's substitute a variable u for 2^x.
So, let u = 2^x.
Now we can rewrite the equation in terms of u:
2(u^2) - 5u + 2 = 0
This is a quadratic equation, which we can solve by factoring or using the quadratic formula.
Let's try factoring:
(2u - 1)(u - 2) = 0
To find the values of u, we set each factor equal to zero:
2u - 1 = 0 or u - 2 = 0
Solving these equations will give us the values of u:
2u = 1 or u = 2
u = 1/2 or u = 2
Now, we substitute back the value of u in terms of x:
2^x = 1/2 or 2^x = 2
To solve for x, we take the logarithm of both sides of each equation:
x = log(1/2) / log(2) or x = log(2) / log(2)
Calculating these values will give us:
x ≈ -0.693 or x = 1
So, the solutions to the equation 2(2^(2x))-5(2^x)+2=0 are x ≈ -0.693 and x = 1.