Solve for x log base 2 of 2x+ log base 2 of x-7 = log base 2 of 4x

$\log_{2} (2 x) + \log_{2} (x - 7) = \log_{2} (4 x)$

Simplify the left side.

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

$\log_{2} (2 x (x - 7)) = \log_{2} (4 x)$

Apply the distributive property.

$\log_{2} (2 x \cdot x + 2 x \cdot - 7) = \log_{2} (4 x)$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$\log_{2} (2 (x \cdot x) + 2 x \cdot - 7) = \log_{2} (4 x)$

Multiply $x$ by $x$ .

$\log_{2} (2 x^{2} + 2 x \cdot - 7) = \log_{2} (4 x)$

Multiply $- 7$ by $2$ .

$\log_{2} (2 x^{2} - 14 x) = \log_{2} (4 x)$

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

$2 x^{2} - 14 x = 4 x$

Solve for $x$ .

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

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

$2 x^{2} - 14 x - 4 x = 0$

Subtract $4 x$ from $- 14 x$ .

$2 x^{2} - 18 x = 0$

Factor $2 x$ out of $2 x^{2} - 18 x$ .

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

$2 x (x) - 18 x = 0$

Factor $2 x$ out of $- 18 x$ .

$2 x (x) + 2 x (- 9) = 0$

Factor $2 x$ out of $2 x (x) + 2 x (- 9)$ .

$2 x (x - 9) = 0$

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

$x = 0$

$x - 9 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

The final solution is all the values that make $2 x (x - 9) = 0$ true.

$x = 0, 9$

Exclude the solutions that do not make $\log_{2} (2 x) + \log_{2} (x - 7) = \log_{2} (4 x)$ true.

$x = 9$