Solve for x natural log of x-4 = natural log of x+2+ natural log of 3

$\ln (x - 4) = \ln (x + 2) + \ln (3)$

Simplify the right side.

Simplify $\ln (x + 2) + \ln (3)$ .

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

$\ln (x - 4) = \ln ((x + 2) \cdot 3)$

Apply the distributive property.

$\ln (x - 4) = \ln (x \cdot 3 + 2 \cdot 3)$

Simplify the expression.

Move $3$ to the left of $x$ .

$\ln (x - 4) = \ln (3 \cdot x + 2 \cdot 3)$

Multiply $2$ by $3$ .

$\ln (x - 4) = \ln (3 x + 6)$

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

$x - 4 = 3 x + 6$

Solve for $x$ .

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

Subtract $3 x$ from both sides of the equation.

$x - 4 - 3 x = 6$

Subtract $3 x$ from $x$ .

$- 2 x - 4 = 6$

Move all terms not containing $x$ to the right side of the equation.

Add $4$ to both sides of the equation.

$- 2 x = 6 + 4$

Add $6$ and $4$ .

$- 2 x = 10$

Divide each term in $- 2 x = 10$ by $- 2$ and simplify.

Divide each term in $- 2 x = 10$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{10}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{10}{- 2}$

Divide $x$ by $1$ .

$x = \frac{10}{- 2}$

Simplify the right side.

Divide $10$ by $- 2$ .

$x = - 5$

Exclude the solutions that do not make $\ln (x - 4) = \ln (x + 2) + \ln (3)$ true.

$No solution$