Answer:
[tex]x=\pm \sqrt{17}[/tex]
Step-by-step explanation:
Given : [tex]ln( {x}^{2} - 16) = 0[/tex]
To Find: x
Solution:
[tex]ln( {x}^{2} - 16) = 0[/tex]
Take antilogarithm of both sides to base e.
[tex]{e}^{ ln( {x}^{2} - 16) } = {e}^{0}[/tex]
[tex]{x}^{2} - 16 = 1[/tex]
Now, Group like terms
[tex]{x}^{2} = 1 + 16[/tex]
[tex]{x}^{2} = 17[/tex]
[tex]x=\pm \sqrt{17}[/tex]
Thus the solution to [tex]ln( {x}^{2} - 16) = 0[/tex] is [tex]x=\pm \sqrt{17}[/tex]
