\( e^{2 x}-6 e^{x}+5=0 \) The solution set expressed in terms of logarithms is \( \{0 \), In 5\( \} \). 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 \( \left\{\begin{array}{l}\text { Solve } \\ \text { (Use a comma to seprate answers as needed. Round to two decimal places as needed.) }\end{array}\right. \)

We start with the equation \[ e^{2x} - 6e^x + 5 = 0. \] **Step 1.** Let \( u = e^x \). Then the equation becomes \[ u^2 - 6u + 5 = 0. \] **Step 2.** Factor the quadratic: \[ (u-1)(u-5)=0. \] Thus, \( u = 1 \) or \( u = 5 \). **Step 3.** Substitute back \( e^x \) for \( u \): - For \( u = 1 \): \[ e^x = 1 \quad \Longrightarrow \quad x = \ln 1 = 0. \] - For \( u = 5 \): \[ e^x = 5 \quad \Longrightarrow \quad x = \ln 5. \] So, the solution set expressed in terms of logarithms is \[ \{0,\, \ln 5\}. \] **Step 4.** Now, we use a calculator to obtain a decimal approximation for the solution \( x = \ln 5 \): \[ \ln 5 \approx 1.61. \] Thus, the solution set, rounded to two decimal places, is \[ \{0,\, 1.61\}. \]