Question

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

Answer 1

Answer:

$x=\pm \sqrt{17}$

Step-by-step explanation:

Given : $ln( {x}^{2} - 16) = 0$

To Find: x

Solution:

$ln( {x}^{2} - 16) = 0$

Take antilogarithm of both sides to base e.  

${e}^{ ln( {x}^{2} - 16) } = {e}^{0}$

${x}^{2} - 16 = 1$

Now, Group like terms

${x}^{2} = 1 + 16$

${x}^{2} = 17$

$x=\pm \sqrt{17}$

Thus the solution to   $ln( {x}^{2} - 16) = 0$  is    $x=\pm \sqrt{17}$

Answer 2

ANSWER

$x = \pm \sqrt{17}$
EXPLANATION

The given equation is

$ln( {x}^{2} - 16) = 0$

Take antilogarithm of both sides to base e.

${e}^{ ln( {x}^{2} - 16) } = {e}^{0}$

This will simplify to

${x}^{2} - 16 = 1$

Group like terms to obtain,

${x}^{2} = 1 + 16$

${x}^{2} = 17$

Take square root of both sides to get,

$x = \pm \sqrt{17}$