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What is the solution to [tex]\ln \left(x^2-16\right)=0[/tex]?

A. [tex]x= \pm \sqrt{17}[/tex]

B. [tex]x= \pm 4[/tex]

C. [tex]x= \pm \sqrt{32}[/tex]


Sagot :

Sure, let’s solve the equation [tex]\(\ln \left(x^2-16\right)=0\)[/tex] step-by-step.

1. Understand the given equation:
[tex]\[\ln \left(x^2 - 16\right) = 0\][/tex]

2. Eliminate the natural logarithm by exponentiating both sides to remove the [tex]\(\ln\)[/tex]:
[tex]\[e^{\ln \left(x^2 - 16\right)} = e^0\][/tex]
Since [tex]\(e^{\ln y} = y\)[/tex] for any [tex]\(y > 0\)[/tex], this simplifies to:
[tex]\[x^2 - 16 = 1\][/tex]

3. Solve the resulting algebraic equation:
[tex]\[x^2 - 16 = 1\][/tex]
Add 16 to both sides:
[tex]\[x^2 = 17\][/tex]

4. Solve for [tex]\(x\)[/tex] by taking the square root of both sides:
[tex]\[x = \pm \sqrt{17}\][/tex]

Therefore, the solutions to the equation [tex]\(\ln \left(x^2 - 16\right) = 0\)[/tex] are:
[tex]\[x = \sqrt{17} \quad \text{and} \quad x = -\sqrt{17}\][/tex]

Thus, the correct option from the given choices is:
[tex]\[x = \pm \sqrt{17}\][/tex]
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