Question

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

Answer 1

`x=\pm √(17)`

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 √(17)`

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

Answer 2

ANSWER


`x = \pm √(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 √(17)`