Solve the following exponential equation. Express the solution set in terms of natural logarithms or common loganthms. Then, us $$ e^{2 x}-4 e^{x}+3=0 $$ The solution set expressed in terms of logarithms is $$ \{0 $$, In 3$$ \} $$. Use a comma to separate answers as needed. Simplify your answer. Use integers or fractions for any numbers in the expression Now use a calculator to obtain a decimal approximation for the solution. The solution set is $$ \left\{\begin{array}{l}\text { Ne answers as needed. Round to two decimal places as needed.) } \ \text { (Use a comma to separate answ }\end{array}\right. $$

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UpStudy AI · Accepted Answer

We start with the given equation  
$$
e^{2x} - 4e^x + 3 = 0.
$$
Introduce the substitution  
$$
u = e^x.
$$
Then the equation becomes  
$$
u^2 - 4u + 3 = 0.
$$

Factor the quadratic:  
$$
(u-1)(u-3)=0.
$$
This gives the solutions  
$$
u = 1 \quad \text{or} \quad u = 3.
$$

Recall that $$ u = e^x $$. Thus, for $$ u=1 $$ we have  
$$
e^x = 1 \quad \Longrightarrow \quad x = \ln(1) = 0.
$$
For $$ u=3 $$ we have  
$$
e^x = 3 \quad \Longrightarrow \quad x = \ln(3).
$$

So the solution set expressed in terms of logarithms is  
$$
\{0,\, \ln(3)\}.
$$

Using a calculator, $$ \ln(3) $$ is approximately  
$$
\ln(3) \approx 1.10.
$$

Thus, the solution set with decimal approximations is  
$$
\{0,\, 1.10\}.
$$
The solution set is $$ \{0, \ln(3)\} $$, which is approximately $$ \{0, 1.10\} $$.

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