Solve ln (x^2 - 16) = 0

Evaluate

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

Find the domain

More Steps

ln (x^{2} - 16) = 0, x \in (- \infty, - 4) \cup (4, + \infty)

Convert the logarithm into exponential form using the fact that lo g_{a} x = b is equal to x = a^{b}

x^{2} - 16 = e^{0}

Evaluate the power

x^{2} - 16 = 1

Move the constant to the right-hand side and change its sign

x^{2} = 1 + 16

Add the numbers

x^{2} = 17

Take the root of both sides of the equation and remember to use both positive and negative roots

x = \pm 17

Separate the equation into 2 possible cases

x = 17 x = - 17

Check if the solution is in the defined range

x = 17 x = - 17, x \in (- \infty, - 4) \cup (4, + \infty)

Find the intersection of the solution and the defined range

x = 17 x = - 17

Solution

x_{1} = - 17, x_{2} = 17

Alternative Form

x_{1} \approx - 4.123106, x_{2} \approx 4.123106

Solve the equation