https://file.aiquickdraw.com/mathbot/file/172616064865140u3m9uo.jpg

The equation you have provided is:

\[
(\ln x)^2 - 4 \ln x + 3 = 0
\]

This is a quadratic equation in terms of \(\ln x\). Let's solve it step-by-step.

### Step 1: Substitution
Let \( u = \ln x \). This transforms the equation into:

\[
u^2 - 4u + 3 = 0
\]

### Step 2: Solve the quadratic equation
We can solve this quadratic equation using factorization or the quadratic formula. The factorization method gives:

\[
(u - 3)(u - 1) = 0
\]

So, \( u = 3 \) or \( u = 1 \).

### Step 3: Back-substitution
Recall that \( u = \ln x \). Thus, we have:

1. \( \ln x = 3 \)
2. \( \ln x = 1 \)

### Step 4: Solve for \( x \)
Now, solve for \( x \) by exponentiating both sides:

1. \( \ln x = 3 \) implies \( x = e^3 \)
2. \( \ln x = 1 \) implies \( x = e^1 = e \)

### Final Answer:
The solutions are:
\[
x = e^3 \quad \text{or} \quad x = e
\]

Let me know if you would like further clarification or more details!

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Here are some related questions to deepen your understanding:

1. What is the general form of a quadratic equation, and how do you solve it?
2. How does logarithmic differentiation work, and when is it useful?
3. Can you solve a logarithmic equation where the base is not \(e\)?
4. What are the properties of logarithms that are commonly used in simplifying expressions?
5. How can you solve more complex equations involving both logarithms and exponents?

**Tip:** Always remember that the logarithmic function \( \ln x \) is only defined for positive values of \( x \).

Learn how to solve the logarithmic quadratic equation (ln(x))^2 - 4ln(x) + 3 = 0 using algebraic methods and logarithmic properties. This guide walks you through step-by-step solutions suitable for Grades 9-12.

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