To solve the equation [tex]\(2^x + 2^{1-x} = 3\)[/tex], we can proceed with the following steps:
1. Introduce a substitution:
Let [tex]\(y = 2^x\)[/tex]. Thus, [tex]\(2^{1-x}\)[/tex] can be rewritten using properties of exponents:
[tex]\[
2^{1-x} = \frac{2}{2^x} = \frac{2}{y}
\][/tex]
So, the original equation [tex]\(2^x + 2^{1-x} = 3\)[/tex] becomes:
[tex]\[
y + \frac{2}{y} = 3
\][/tex]
2. Multiply through by [tex]\(y\)[/tex] to eliminate the fraction:
[tex]\[
y^2 + 2 = 3y
\][/tex]
3. Move all terms to one side to set up for factoring:
[tex]\[
y^2 - 3y + 2 = 0
\][/tex]
4. Factor the quadratic equation:
We need to find factors of [tex]\(2\)[/tex] that add up to [tex]\(-3\)[/tex]. The correct factors are [tex]\(-1\)[/tex] and [tex]\(-2\)[/tex]:
[tex]\[
(y - 1)(y - 2) = 0
\][/tex]
5. Solve the factored equation for [tex]\(y\)[/tex]:
[tex]\[
y - 1 = 0 \quad \text{or} \quad y - 2 = 0
\][/tex]
So, [tex]\(y = 1\)[/tex] or [tex]\(y = 2\)[/tex].
6. Recall the substitution [tex]\(y = 2^x\)[/tex]:
[tex]\[
2^x = 1 \quad \text{or} \quad 2^x = 2
\][/tex]
7. Solve for [tex]\(x\)[/tex] in each case:
- If [tex]\(2^x = 1\)[/tex]:
[tex]\[
2^x = 2^0 \rightarrow x = 0
\][/tex]
- If [tex]\(2^x = 2\)[/tex]:
[tex]\[
2^x = 2^1 \rightarrow x = 1
\][/tex]
Therefore, the solutions to the equation [tex]\(2^x + 2^{1-x} = 3\)[/tex] are:
[tex]\[
x = 0 \quad \text{or} \quad x = 1
\][/tex]
