Ork 3_Chapter 3 (Course Work \( 1 \leftarrow \) Solve the following exponential equation. Express the solution set in terms of natural logarithms or common logarithms. Then, us \( e^{2 x}-6 e^{x}+5=0 \) The solution set expressed in terms of logarithms is \( 58,3.4 .43 \) 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 \( e^{2x} - 6e^{x} + 5 = 0 \), we start by substituting \( y = e^{x} \). This transforms the equation into a quadratic form: \[ y^2 - 6y + 5 = 0. \] Factoring gives us \( (y - 5)(y - 1) = 0 \), so \( y = 5 \) or \( y = 1 \). Now, revert back to \( e^{x} \): 1. For \( y = 5 \): \( e^{x} = 5 \) leads to \( x = \ln(5) \). 2. For \( y = 1 \): \( e^{x} = 1 \) gives \( x = \ln(1) = 0 \). Thus, the solution set expressed in terms of logarithms is \( \ln(5), 0 \). Don't forget to check your work; substituting back into the original equation helps confirm that your solutions are correct! Happy solving!