Solve the following exponential equation. Express the solution set in terms of natural logarithms or common logarithms. Then, use a \[ e^{2 x}-3 e^{x}+2=0 \] The solution set expressed in terms of logarithms is \( \{0, \ln 2\} \). Use a comma to separate answers as needed. Simplify your answer. Use integers or fractions for any numbers in the expression. Use Now use a calculator to obtain a decimal approximation for the solution. The solution set is \( \left\{\begin{array}{l}\text {. }\end{array}\right. \) (Use a comma to separate answers as needed. Round to two decimal places as needed.)

To solve the exponential equation \( e^{2x} - 3e^x + 2 = 0 \), we can make a substitution. Let \( y = e^x \). Then, the equation becomes \( y^2 - 3y + 2 = 0 \). Factoring gives us \( (y - 1)(y - 2) = 0 \), so \( y = 1 \) or \( y = 2 \). Re-substituting for \( x \), we have \( e^x = 1 \) (which leads to \( x = 0 \)) and \( e^x = 2 \) (which gives \( x = \ln 2 \)). Hence, the solution set in terms of logarithms is \( \{0, \ln 2\} \). Now, let's calculate the decimal approximations for those values. The approximate value for \( 0 \) remains \( 0.00 \). For \( \ln 2 \), using a calculator, we find that it is approximately \( 0.69 \). Therefore, the solution set in decimal form is \( \{0.00, 0.69\} \).