Trigonometry Examples

$\ln (x^{2} - 5) = \ln (4 x)$

Step 1

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$x^{2} - 5 = 4 x$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Subtract $4 x$ from both sides of the equation.

$x^{2} - 5 - 4 x = 0$

Step 2.2

Factor $x^{2} - 5 - 4 x$ using the AC method.

Tap for more steps...

Step 2.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $- 4$ .

$- 5, 1$

Step 2.2.2

Write the factored form using these integers.

$(x - 5) (x + 1) = 0$

Step 2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 5 = 0$

$x + 1 = 0$

Step 2.4

Set $x - 5$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.4.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 2.4.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.5

Set $x + 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.5.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 2.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.6

The final solution is all the values that make $(x - 5) (x + 1) = 0$ true.

$x = 5, - 1$

Step 3

Exclude the solutions that do not make $\ln (x^{2} - 5) = \ln (4 x)$ true.

$x = 5$