Solve the following exponential equation. Express the solution set in terms of natural logarithms or common logarithms. Then, use \[ e^{2 x}-6 e^{x}+5=0 \] The solution set expressed in terms of logarithms is \( \square \) ]. Use a comma to separate answers as needed. Simplify your answer. Use integers or fractions for any numbers in the expression.

To solve the equation \( e^{2x} - 6e^{x} + 5 = 0 \), let’s make a substitution. Let \( y = e^{x} \). This transforms the equation into a standard quadratic form: \[ y^2 - 6y + 5 = 0 \] Now we can factor this equation: \[ (y - 1)(y - 5) = 0 \] Setting each factor to zero gives us: \[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \] \[ y - 5 = 0 \quad \Rightarrow \quad y = 5 \] Now we revert to our original variable using \( y = e^{x} \): 1. From \( e^{x} = 1 \): \[ x = \ln(1) = 0 \] 2. From \( e^{x} = 5 \): \[ x = \ln(5) \] Thus, the solution set expressed in terms of logarithms is: \[ \{ 0, \ln(5) \} \]