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!