Question

what is the solution to $ln(x^2 - 16) = 0$?
$x = \pm \sqrt{17}$
$x = \pm 4$
$x = \pm \sqrt{32}$
done

Explanation:

Step1: Recall the property of natural logarithm

We know that if \(\ln(a)=b\), then \(a = e^{b}\). In this equation, \(\ln(x^{2}-16)=0\), so by the property of natural logarithm, we have \(x^{2}-16=e^{0}\).

Step2: Simplify \(e^{0}\)

Since any non - zero number to the power of \(0\) is \(1\), \(e^{0} = 1\). So the equation becomes \(x^{2}-16 = 1\).

Step3: Solve for \(x^{2}\)

Add \(16\) to both sides of the equation \(x^{2}-16=1\). We get \(x^{2}=1 + 16=17\).

Step4: Solve for \(x\)

Take the square root of both sides. Since if \(x^{2}=a\) (\(a\geq0\)), then \(x=\pm\sqrt{a}\), for \(x^{2}=17\), we have \(x = \pm\sqrt{17}\).

Answer:

\(x=\pm\sqrt{17}\) (corresponding to the first option \(x = \pm\sqrt{17}\))