Answer :
To solve the equation \( 2^{x-2} + 2^{3-x} = 3 \), follow these steps:
1. Introduce a substitution to simplify the equation:
Let \( y = 2^x \). Therefore, \( 2^{x-2} = \frac{y}{4} \) and \( 2^{3-x} = \frac{8}{y} \).
2. Rewrite the equation with the substitution:
[tex]\[ \frac{y}{4} + \frac{8}{y} = 3 \][/tex]
3. Multiply through by \( y \) to clear the fractions:
[tex]\[ y \cdot \frac{y}{4} + y \cdot \frac{8}{y} = 3y \][/tex]
[tex]\[ \frac{y^2}{4} + 8 = 3y \][/tex]
4. Multiply all terms by 4 to eliminate the denominator:
[tex]\[ y^2 + 32 = 12y \][/tex]
5. Rearrange into standard quadratic form:
[tex]\[ y^2 - 12y + 32 = 0 \][/tex]
6. Solve the quadratic equation:
Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -12 \), and \( c = 32 \).
[tex]\[ y = \frac{12 \pm \sqrt{144 - 128}}{2} \][/tex]
[tex]\[ y = \frac{12 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ y = \frac{12 \pm 4}{2} \][/tex]
This gives us the solutions:
[tex]\[ y = \frac{16}{2} = 8 \quad \text{and} \quad y = \frac{8}{2} = 4 \][/tex]
7. Back-substitute to find \( x \):
Recall that \( y = 2^x \). Now solve for \( x \).
For \( y = 8 \):
[tex]\[ 8 = 2^x \implies 2^x = 2^3 \implies x = 3 \][/tex]
For \( y = 4 \):
[tex]\[ 4 = 2^x \implies 2^x = 2^2 \implies x = 2 \][/tex]
Therefore, the solutions to the equation [tex]\( 2^{x-2} + 2^{3-x} = 3 \)[/tex] are [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex].
1. Introduce a substitution to simplify the equation:
Let \( y = 2^x \). Therefore, \( 2^{x-2} = \frac{y}{4} \) and \( 2^{3-x} = \frac{8}{y} \).
2. Rewrite the equation with the substitution:
[tex]\[ \frac{y}{4} + \frac{8}{y} = 3 \][/tex]
3. Multiply through by \( y \) to clear the fractions:
[tex]\[ y \cdot \frac{y}{4} + y \cdot \frac{8}{y} = 3y \][/tex]
[tex]\[ \frac{y^2}{4} + 8 = 3y \][/tex]
4. Multiply all terms by 4 to eliminate the denominator:
[tex]\[ y^2 + 32 = 12y \][/tex]
5. Rearrange into standard quadratic form:
[tex]\[ y^2 - 12y + 32 = 0 \][/tex]
6. Solve the quadratic equation:
Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -12 \), and \( c = 32 \).
[tex]\[ y = \frac{12 \pm \sqrt{144 - 128}}{2} \][/tex]
[tex]\[ y = \frac{12 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ y = \frac{12 \pm 4}{2} \][/tex]
This gives us the solutions:
[tex]\[ y = \frac{16}{2} = 8 \quad \text{and} \quad y = \frac{8}{2} = 4 \][/tex]
7. Back-substitute to find \( x \):
Recall that \( y = 2^x \). Now solve for \( x \).
For \( y = 8 \):
[tex]\[ 8 = 2^x \implies 2^x = 2^3 \implies x = 3 \][/tex]
For \( y = 4 \):
[tex]\[ 4 = 2^x \implies 2^x = 2^2 \implies x = 2 \][/tex]
Therefore, the solutions to the equation [tex]\( 2^{x-2} + 2^{3-x} = 3 \)[/tex] are [tex]\( x = 2 \)[/tex] and [tex]\( x = 3 \)[/tex].
