To solve the equation 

\[
\ln(x^2 - 1) = 3,
\]

we start by rewriting it in exponential form to eliminate the natural logarithm. Recall that if \(\ln a = b\), then \(a = e^b\). Applying this property:

\[
x^2 - 1 = e^3.
\]

Next, calculate \(e^3\):

\[
x^2 - 1 = e^3 \approx 20.0855.
\]

Now, solve for \(x^2\):

\[
x^2 = e^3 + 1 \approx 20.0855 + 1 = 21.0855.
\]

Take the square root of both sides to find \(x\):

\[
x = \pm\sqrt{21.0855}.
\]

Calculating the square root:

\[
x \approx \pm 4.592.
\]

Therefore, the solutions are:

\[
x \approx 4.592 \quad \text{and} \quad x \approx -4.592.
\]

### Do you want more details or have any questions?

Here are some related questions you might find useful:
1. How would the solution change if the equation was \(\ln(x^2 + 1) = 3\)?
2. What if the equation were \(\ln(2x^2 - 1) = 3\)?
3. How do you solve \(\ln(x - 1) = 3\)?
4. What is the domain of the function \(\ln(x^2 - 1)\)?
5. How would you handle a logarithmic equation with a different base?

**Tip:** Always check the domain when dealing with logarithmic functions to ensure that the argument is positive!

Discover how to solve the logarithmic equation ln(x^2 - 1) = 3 by converting it into exponential form and using key algebraic techniques. This step-by-step guide explains how to handle logarithmic and quadratic expressions, including calculating e^3 and finding the square root. Suitable for students in Grades 10-12.

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