Sure, let's solve the equation [tex]\( 2^{a-2} + 2^{3-a} = 3 \)[/tex] step-by-step.
### Step 1: Simplify the Equation
First, we can use the property of exponents [tex]\( 2^{a-2} = 2^a \cdot 2^{-2} = \frac{2^a}{4} \)[/tex] and [tex]\( 2^{3-a} = \frac{2^3}{2^a} = \frac{8}{2^a} \)[/tex].
This transforms the equation:
[tex]\[ \frac{2^a}{4} + \frac{8}{2^a} = 3 \][/tex]
### Step 2: Clear the Fractions
To clear the fractions, multiply both sides of the equation by [tex]\( 4 \cdot 2^a \)[/tex]:
[tex]\[ 4 \cdot 2^a \left( \frac{2^a}{4} + \frac{8}{2^a} \right) = 4 \cdot 2^a \cdot 3 \][/tex]
This simplifies to:
[tex]\[ 2^a \cdot 2^a + 4 \cdot 8 = 12 \cdot 2^a \][/tex]
or:
[tex]\[ (2^a)^2 + 32 = 12 \cdot 2^a \][/tex]
### Step 3: Let [tex]\( x = 2^a \)[/tex]
Substitute [tex]\( x \)[/tex] for [tex]\( 2^a \)[/tex]:
[tex]\[ x^2 + 32 = 12x \][/tex]
### Step 4: Solve the Quadratic Equation
Rearrange to form a standard quadratic equation:
[tex]\[ x^2 - 12x + 32 = 0 \][/tex]
Now, solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = 32 \)[/tex]:
[tex]\[ x = \frac{12 \pm \sqrt{144 - 128}}{2} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ x = \frac{12 \pm 4}{2} \][/tex]
So, we have two solutions for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{12 + 4}{2} = 8 \][/tex]
[tex]\[ x = \frac{12 - 4}{2} = 4 \][/tex]
### Step 5: Solve for [tex]\( a \)[/tex]
Recall that [tex]\( x = 2^a \)[/tex]:
- For [tex]\( x = 8 \)[/tex]:
[tex]\[ 2^a = 8 \][/tex]
[tex]\[ 2^a = 2^3 \][/tex]
[tex]\[ a = 3 \][/tex]
- For [tex]\( x = 4 \)[/tex]:
[tex]\[ 2^a = 4 \][/tex]
[tex]\[ 2^a = 2^2 \][/tex]
[tex]\[ a = 2 \][/tex]
### Step 6: Verify the Solutions
Let's verify these solutions by substituting them back into the original equation.
- For [tex]\( a = 3 \)[/tex]:
[tex]\[ 2^{3-2} + 2^{3-3} = 2^1 + 2^0 = 2 + 1 = 3 \][/tex]
- For [tex]\( a = 2 \)[/tex]:
[tex]\[ 2^{2-2} + 2^{3-2} = 2^0 + 2^1 = 1 + 2 = 3 \][/tex]
Both are correct.
### Final Answer
The solutions are:
[tex]\[ a = 2 \text{ and } a = 3 \][/tex]
