To solve the equation [tex]\(3^{2x} - 4 \times 3^x + 3 = 0\)[/tex], we can use substitution to simplify the problem into a more familiar form, specifically a quadratic equation. Here are the detailed steps:
1. Substitution:
Let [tex]\(y = 3^x\)[/tex]. This implies that [tex]\(y^2 = 3^{2x}\)[/tex]. Using this substitution, the original equation becomes:
[tex]\[
y^2 - 4y + 3 = 0
\][/tex]
2. Quadratic Equation:
Now, solve the quadratic equation [tex]\(y^2 - 4y + 3 = 0\)[/tex]. This can be done using the quadratic formula, [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = 3\)[/tex].
3. Apply Quadratic Formula:
[tex]\[
y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}
\][/tex]
[tex]\[
y = \frac{4 \pm \sqrt{16 - 12}}{2}
\][/tex]
[tex]\[
y = \frac{4 \pm \sqrt{4}}{2}
\][/tex]
[tex]\[
y = \frac{4 \pm 2}{2}
\][/tex]
4. Solve for y:
There are two solutions for [tex]\(y\)[/tex]:
[tex]\[
y = \frac{4 + 2}{2} = 3
\][/tex]
[tex]\[
y = \frac{4 - 2}{2} = 1
\][/tex]
5. Back-Substitute:
Recall that [tex]\(y = 3^x\)[/tex]. We now solve for [tex]\(x\)[/tex] using the solutions [tex]\(y = 3\)[/tex] and [tex]\(y = 1\)[/tex]:
- For [tex]\(y = 3\)[/tex]:
[tex]\[
3^x = 3 \implies x = 1
\][/tex]
- For [tex]\(y = 1\)[/tex]:
[tex]\[
3^x = 1 \implies x = 0
\][/tex]
(since [tex]\(3^0 = 1\)[/tex])
6. Solution:
The solutions to the original equation [tex]\(3^{2x} - 4 \times 3^x + 3 = 0\)[/tex] are:
[tex]\[
x = 1 \quad \text{and} \quad x = 0
\][/tex]
Thus, the complete set of solutions is [tex]\( \{0, 1\} \)[/tex].
