Solve for x log of x+ log of x-1 = log of 4x

$\log (x) + \log (x - 1) = \log (4 x)$

Simplify the left side.

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

$\log (x (x - 1)) = \log (4 x)$

Simplify by multiplying through.

Apply the distributive property.

$\log (x \cdot x + x \cdot - 1) = \log (4 x)$

Simplify the expression.

Multiply $x$ by $x$ .

$\log (x^{2} + x \cdot - 1) = \log (4 x)$

Move $- 1$ to the left of $x$ .

$\log (x^{2} - 1 \cdot x) = \log (4 x)$

Rewrite $- 1 x$ as $- x$ .

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

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

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

Solve for $x$ .

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

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

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

Subtract $4 x$ from $- x$ .

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

Factor $x$ out of $x^{2} - 5 x$ .

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

$x \cdot x - 5 x = 0$

Factor $x$ out of $- 5 x$ .

$x \cdot x + x \cdot - 5 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 5$ .

$x (x - 5) = 0$

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

$x = 0$

$x - 5 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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

$x = 0, 5$

Exclude the solutions that do not make $\log (x) + \log (x - 1) = \log (4 x)$ true.

$x = 5$