Question

Solve the following exponential equation. Express the solution set in terms of natural logarithms or common logarithms.​ Then, use a calculator to obtain a decimal approximation for the solution.
e Superscript 2 x Baseline minus 3 e Superscript x Baseline plus 2 equals 0
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Part 1
The solution set expressed in terms of logarithms is StartSet nothing EndSet
  
.
Use a comma to separate answers as needed. Simplify your answer. Use integers or fractions for any numbers in the expression. Use ln for natural logarithm and log for common logarithm.
Part 2
Now use a calculator to obtain a decimal approximation for the solution.
The solution set is StartSet nothing EndSet .
  

​(Use a comma to separate answers as needed. Round to two decimal places as​ needed.)

Answer 1

To solve the given exponential equation \( e^{2x} - 3e^x + 2 = 0 \), we can start by making a substitution to simplify it. Let \( y = e^x \). Then, the equation becomes:

\[
y^2 - 3y + 2 = 0
\]

Now, we can factor this quadratic equation:

\[
(y - 1)(y - 2) = 0
\]

Setting each factor to zero gives us:

\[
y - 1 = 0 \quad \Rightarrow \quad y = 1
\]
\[
y - 2 = 0 \quad \Rightarrow \quad y = 2
\]

Now we substitute back for \( y \):

1. For \( y = 1 \):
\[
e^x = 1 \quad \Rightarrow \quad x = \ln(1) = 0
\]

2. For \( y = 2 \):
\[
e^x = 2 \quad \Rightarrow \quad x = \ln(2)
\]

Now we can express the solution set in terms of logarithms:
- The solution set expressed in terms of logarithms is:
\[
\{0, \ln(2)\}
\]

### Part 2

Now, we will use a calculator to obtain a decimal approximation for the solutions.

1. The decimal approximation for \( x = 0 \) is:
\[
0 \approx 0.00
\]

2. The decimal approximation for \( x = \ln(2) \) is approximately:
\[
\ln(2) \approx 0.6931
\]

Rounding to two decimal places, we have:
\[
\ln(2) \approx 0.69
\]

Thus, the solution set expressed in decimal form is:
\[
\{0.00, 0.69\}
\]

### Summary:
- Part 1 answer: \(\{0, \ln(2)\}\)
- Part 2 answer: \(\{0.00, 0.69\}\)