The equation
2^2x−5×2^x+6=0
has two solutions. The smaller is ? and the larger is ?.
this is just a quadratic in 2^x, since 2^2x = (2^x)^2
So, it factors into
(2^x-3)(2^x-2) = 0
So, 2^x = 2 or 3
That means x = 1 or log2(3)
To find the solutions of the equation 2^(2x) - 5 * 2^x + 6 = 0, we can use a substitution technique. Let's make a substitution to simplify the equation.
Let's substitute y = 2^x. Now our equation becomes y^2 - 5y + 6 = 0.
To solve this quadratic equation, we can factor it. Factoring the equation gives us (y - 2)(y - 3) = 0.
Setting each factor equal to zero, we have two possible solutions:
y - 2 = 0, which gives y = 2,
y - 3 = 0, which gives y = 3.
Now, we need to find the corresponding values of x. We can use the substitution y = 2^x to find the values of x.
For y = 2, we have 2^x = 2. Solving this gives us x = 1.
For y = 3, we have 2^x = 3. This equation doesn't have an exact solution using ordinary algebra. However, we can use logarithms to approximate the value of x.
Taking the logarithm of both sides of the equation 2^x = 3, we get log(2^x) = log(3).
Using the logarithmic property, we can rewrite it as x log(2) = log(3).
Dividing both sides of the equation by log(2), we get x = log(3) / log(2).
Using a calculator, we can evaluate log(3) and log(2) to approximate x.
The approximate value of x is x ≈ 1.585.
So, the smaller solution is x = 1, and the larger solution is x ≈ 1.585.