To solve the equation [tex]\( 2^{x-2} + 2^{3-x} = 3 \)[/tex], let's proceed with a step-by-step algebraic approach.
1. Substitution: Let us denote [tex]\( y = 2^{x-2} \)[/tex].
2. Rewriting in terms of [tex]\( y \)[/tex]:
- Note that [tex]\( 2^{x-2} = y \)[/tex].
- Recall that [tex]\( 2^{3-x} \)[/tex] can be rewritten in terms of [tex]\( 2^{x-2} \)[/tex]. Specifically, since [tex]\( 2^{3-x} = 2^{3-(x-2+2)} = 2^{1-(x-2)}\)[/tex], we get [tex]\( 2^{3-x} = \frac{8}{2^{x-2}} = \frac{8}{y} \)[/tex].
Therefore, our equation transforms into:
[tex]\[
y + \frac{8}{y} = 3
\][/tex]
3. Form a quadratic equation: Multiply both sides by [tex]\( y \)[/tex] to clear the fraction:
[tex]\[
y^2 + 8 = 3y
\][/tex]
[tex]\[
y^2 - 3y + 8 = 0
\][/tex]
4. Solve the quadratic equation: To solve the quadratic [tex]\( y^2 - 3y + 8 = 0 \)[/tex], we need to find the roots of the equation. The standard form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = 8 \)[/tex].
The quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] can be used here:
[tex]\[
y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}
\][/tex]
[tex]\[
y = \frac{3 \pm \sqrt{9 - 32}}{2}
\][/tex]
[tex]\[
y = \frac{3 \pm \sqrt{-23}}{2}
\][/tex]
The discriminant ([tex]\( \sqrt{-23} \)[/tex]) is negative, indicating that the quadratic equation has no real solutions.
Since [tex]\( y = 2^{x-2} \)[/tex] must be a positive real number (as it is an exponential expression), and we have found that there are no real solutions for [tex]\( y \)[/tex], we conclude that the original equation [tex]\( 2^{x-2} + 2^{3-x} = 3 \)[/tex] has no real solutions.
Therefore, there are no real values of [tex]\( x \)[/tex] that satisfy the equation [tex]\( 2^{x-2} + 2^{3-x} = 3 \)[/tex].
