What is the solution to ln (x2 - 16) = 0?

ANSWER

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

The given equation is

[tex] ln( {x}^{2} - 16) = 0[/tex]

Take antilogarithm of both sides to base e.

[tex] {e}^{ ln( {x}^{2} - 16) } = {e}^{0} [/tex]

This will simplify to

[tex] {x}^{2} - 16 = 1[/tex]

Group like terms to obtain,

[tex]{x}^{2} = 1 + 16[/tex]

[tex]{x}^{2} = 17[/tex]

Take square root of both sides to get,

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