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!

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 $x$.