To solve the equation [tex]\(4^x - 6 \times 2^{x+1} + 32 = 0\)[/tex], we can follow these detailed steps:
1. Rewrite the given equation using properties of exponents:
[tex]\[ 4^x - 6 \times 2^{x+1} + 32 = 0 \][/tex]
Note that [tex]\(4^x\)[/tex] can be rewritten as [tex]\((2^2)^x\)[/tex], which simplifies to [tex]\((2^x)^2\)[/tex]. Therefore, the equation becomes:
[tex]\[ (2^x)^2 - 6 \times 2^{x+1} + 32 = 0 \][/tex]
2. Introduce a substitution to simplify the equation:
Let [tex]\( y = 2^x \)[/tex]. Thus, the equation transforms into a quadratic form:
[tex]\[ y^2 - 6 \times 2y + 32 = 0 \][/tex]
Simplify the middle term:
[tex]\[ y^2 - 12y + 32 = 0 \][/tex]
3. Solve the quadratic equation:
The quadratic equation [tex]\( y^2 - 12y + 32 = 0 \)[/tex] can be solved using the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\(a = 1\)[/tex], [tex]\(b = -12\)[/tex], and [tex]\(c = 32\)[/tex].
4. Calculate the discriminant:
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values:
[tex]\[ \Delta = (-12)^2 - 4 \times 1 \times 32 \][/tex]
[tex]\[ \Delta = 144 - 128 \][/tex]
[tex]\[ \Delta = 16 \][/tex]
5. Find the roots [tex]\(y_1\)[/tex] and [tex]\(y_2\)[/tex]:
[tex]\[ y_{1,2} = \frac{12 \pm \sqrt{16}}{2 \times 1} \][/tex]
[tex]\[ y_{1,2} = \frac{12 \pm 4}{2} \][/tex]
This gives us the two roots:
[tex]\[ y_1 = \frac{12 + 4}{2} = \frac{16}{2} = 8 \][/tex]
[tex]\[ y_2 = \frac{12 - 4}{2} = \frac{8}{2} = 4 \][/tex]
6. Back-substitute [tex]\(y = 2^x\)[/tex]:
We have [tex]\( y = 2^x \)[/tex]. Thus, we get two possible equations:
[tex]\[ 2^x = 8 \][/tex]
[tex]\[ 2^x = 4 \][/tex]
7. Solve for [tex]\(x\)[/tex] using the logarithm base 2:
[tex]\[ 2^x = 8 \implies x = \log_2 8 \][/tex]
[tex]\[ 8 = 2^3 \implies x = 3 \][/tex]
[tex]\[ 2^x = 4 \implies x = \log_2 4 \][/tex]
[tex]\[ 4 = 2^2 \implies x = 2 \][/tex]
Therefore, the solutions to the equation [tex]\(4^x - 6 \times 2^{x+1} + 32 = 0\)[/tex] are:
[tex]\[ x = 3 \quad \text{and} \quad x = 2 \][/tex]
