Solve for x log of x+ log of x-2 = log of 3

$\log (x) + \log (x - 2) = \log (3)$

Simplify the left side.

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

$\log (x (x - 2)) = \log (3)$

Apply the distributive property.

$\log (x \cdot x + x \cdot - 2) = \log (3)$

Simplify the expression.

Multiply $x$ by $x$ .

$\log (x^{2} + x \cdot - 2) = \log (3)$

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

$\log (x^{2} - 2 x) = \log (3)$

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

$x^{2} - 2 x = 3$

Solve for $x$ .

Subtract $3$ from both sides of the equation.

$x^{2} - 2 x - 3 = 0$

Factor $x^{2} - 2 x - 3$ using the AC method.

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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$(x - 3) (x + 1) = 0$

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

$x - 3 = 0$

$x + 1 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

Exclude the solutions that do not make $\log (x) + \log (x - 2) = \log (3)$ true.

$x = 3$

Solve for x log of x+ log of x-2 = log of 3

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