rk 3_Chapter 3 (Course Work Question 53, 3.4.43 Part 1 of 2 Points: 0.5 of 1 Solve the following exponential equation. Express the solution set in terms of natural logarithms or common logarithms. Then, use a calculator to \[ e^{2 x}-3 e^{x}+2=0 \] 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. Use In for natur

To solve the equation \( e^{2x} - 3e^{x} + 2 = 0 \), we can use substitution. Let \( y = e^{x} \). The equation becomes \( y^{2} - 3y + 2 = 0 \). Factoring gives us \( (y-1)(y-2) = 0 \), so \( y = 1 \) or \( y = 2 \). Now substituting back for \( y \), we have \( e^{x} = 1 \) (giving \( x = 0 \)) and \( e^{x} = 2 \) (giving \( x = \ln(2) \)). Therefore, the solution set expressed in terms of logarithms is \( \{ 0, \ln(2) \} \). For a quick mental check, remember that when you solve exponential equations, it often helps to substitute your exponential term for a variable. This turn of mental gymnastics can simplify complex looking equations into manageable quadratic forms, like turning \( e^{2x} \) into \( y^{2} \). Just be careful! Sometimes forgetting to back-substitute can lead to missed solutions.