Precalculus Examples

$\log (x) + \log (x - 2) = \log (5 x)$

Step 1

Simplify the left side.

Tap for more steps...

Step 1.1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log (x (x - 2)) = \log (5 x)$

Step 1.2

Apply the distributive property.

$\log (x \cdot x + x \cdot - 2) = \log (5 x)$

Step 1.3

Simplify the expression.

Tap for more steps...

Step 1.3.1

Multiply $x$ by $x$ .

$\log (x^{2} + x \cdot - 2) = \log (5 x)$

Step 1.3.2

Move $- 2$ to the left of $x$ .

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

Step 2

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

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

Step 3

Solve for $x$ .

Tap for more steps...

Step 3.1

Move all terms containing $x$ to the left side of the equation.

Tap for more steps...

Step 3.1.1

Subtract $5 x$ from both sides of the equation.

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

Step 3.1.2

Subtract $5 x$ from $- 2 x$ .

$x^{2} - 7 x = 0$

Step 3.2

Factor $x$ out of $x^{2} - 7 x$ .

Tap for more steps...

Step 3.2.1

Factor $x$ out of $x^{2}$ .

$x \cdot x - 7 x = 0$

Step 3.2.2

Factor $x$ out of $- 7 x$ .

$x \cdot x + x \cdot - 7 = 0$

Step 3.2.3

Factor $x$ out of $x \cdot x + x \cdot - 7$ .

$x (x - 7) = 0$

Step 3.3

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

$x = 0$

$x - 7 = 0$

Step 3.4

Set $x$ equal to $0$ .

$x = 0$

Step 3.5

Set $x - 7$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 3.5.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 3.5.2

Add $7$ to both sides of the equation.

$x = 7$

Step 3.6

The final solution is all the values that make $x (x - 7) = 0$ true.

$x = 0, 7$

Step 4

Exclude the solutions that do not make $\log (x) + \log (x - 2) = \log (5 x)$ true.

$x = 7$