Certainly! Let's solve the equation [tex]\( 5^{x+1} + 5^{2-x} = 126 \)[/tex] step by step.
### Step 1: Simplify the Equation
Rewrite the terms in a similar base structure to better analyze the equation:
[tex]\[ 5^{x+1} + 5^{2-x} = 126 \][/tex]
Recognize that:
[tex]\[ 5^{x+1} = 5 \cdot 5^x \][/tex]
[tex]\[ 5^{2-x} = 5^2 \cdot 5^{-x} = 25 \cdot 5^{-x} \][/tex]
Thus, the equation becomes:
[tex]\[ 5 \cdot 5^x + 25 \cdot 5^{-x} = 126 \][/tex]
### Step 2: Introducing a New Variable
Let [tex]\( y = 5^x \)[/tex]. This implies that [tex]\( 5^{-x} = \frac{1}{5^x} = \frac{1}{y} \)[/tex].
Substituting [tex]\( y \)[/tex] into the equation, we get:
[tex]\[ 5y + 25 \cdot \frac{1}{y} = 126 \][/tex]
### Step 3: Solve for [tex]\( y \)[/tex]
Multiply through by [tex]\( y \)[/tex] to eliminate the fraction:
[tex]\[ 5y^2 + 25 = 126y \][/tex]
Rearrange the equation into a standard quadratic form:
[tex]\[ 5y^2 - 126y + 25 = 0 \][/tex]
### Step 4: Solve the Quadratic Equation
Use the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 5 \)[/tex], [tex]\( b = -126 \)[/tex], and [tex]\( c = 25 \)[/tex]:
[tex]\[ y = \frac{-(-126) \pm \sqrt{(-126)^2 - 4 \cdot 5 \cdot 25}}{2 \cdot 5} \][/tex]
[tex]\[ y = \frac{126 \pm \sqrt{15876 - 500}}{10} \][/tex]
[tex]\[ y = \frac{126 \pm \sqrt{15376}}{10} \][/tex]
[tex]\[ y = \frac{126 \pm 124}{10} \][/tex]
So we have two potential solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{126 + 124}{10} = \frac{250}{10} = 25 \][/tex]
[tex]\[ y = \frac{126 - 124}{10} = \frac{2}{10} = 0.2 \][/tex]
### Step 5: Back-Substitute to Find [tex]\( x \)[/tex]
Recall that [tex]\( y = 5^x \)[/tex]:
1. If [tex]\( y = 25 \)[/tex]:
[tex]\[ 5^x = 25 \][/tex]
[tex]\[ 5^x = 5^2 \][/tex]
[tex]\[ x = 2 \][/tex]
2. If [tex]\( y = 0.2 \)[/tex]:
[tex]\[ 5^x = 0.2 \][/tex]
[tex]\[ 0.2 = \frac{1}{5} \][/tex]
[tex]\[ 5^x = 5^{-1} \][/tex]
[tex]\[ x = -1 \][/tex]
### Conclusion
The solutions to the equation [tex]\( 5^{x+1} + 5^{2-x} = 126 \)[/tex] are:
[tex]\[ x = 2 \quad \text{or} \quad x = -1 \][/tex]
