Certainly! Let's solve the given exponential equation step-by-step:
We are given:
[tex]\[ 2^x + 2^{-x} = \frac{5}{2} \][/tex]
### Step 1: Substitution
Let us substitute [tex]\( y = 2^x \)[/tex]. Consequently, [tex]\( 2^{-x} = \frac{1}{2^x} = \frac{1}{y} \)[/tex]. The equation becomes:
[tex]\[ y + \frac{1}{y} = \frac{5}{2} \][/tex]
### Step 2: Clear the Fraction
To clear the fraction, multiply every term by [tex]\( y \)[/tex]:
[tex]\[ y^2 + 1 = \frac{5}{2}y \][/tex]
### Step 3: Rearrange into Standard Quadratic Form
Rearrange the equation to get it into the standard quadratic form [tex]\( ax^2 + bx + c = 0 \)[/tex]:
[tex]\[ 2y^2 - 5y + 2 = 0 \][/tex]
### Step 4: Solve the Quadratic Equation
We can solve the quadratic equation using the quadratic formula, [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = -5 \)[/tex], and [tex]\( c = 2 \)[/tex]:
[tex]\[ y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \][/tex]
[tex]\[ y = \frac{5 \pm \sqrt{25 - 16}}{4} \][/tex]
[tex]\[ y = \frac{5 \pm \sqrt{9}}{4} \][/tex]
[tex]\[ y = \frac{5 \pm 3}{4} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{5 + 3}{4} = \frac{8}{4} = 2 \][/tex]
[tex]\[ y = \frac{5 - 3}{4} = \frac{2}{4} = \frac{1}{2} \][/tex]
### Step 5: Back-substitute [tex]\( y = 2^x \)[/tex]
Now, recall that [tex]\( y = 2^x \)[/tex], so we substitute back to find [tex]\( x \)[/tex].
For [tex]\( y = 2 \)[/tex]:
[tex]\[ 2^x = 2 \][/tex]
By inspection, [tex]\( x = 1 \)[/tex].
For [tex]\( y = \frac{1}{2} \)[/tex]:
[tex]\[ 2^x = \frac{1}{2} \][/tex]
[tex]\[ 2^x = 2^{-1} \][/tex]
By inspection, [tex]\( x = -1 \)[/tex].
### Step 6: State the Solutions
So, the solutions to the equation [tex]\( 2^x + 2^{-x} = \frac{5}{2} \)[/tex] are:
[tex]\[ x = -1 \][/tex]
[tex]\[ x = 1 \][/tex]
Therefore, the values of [tex]\( x \)[/tex] that satisfy the given equation are:
[tex]\[ x = -1 \quad \text{and} \quad x = 1 \][/tex]
