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

$\log (3 x) = \log (4) + \log (x - 3)$

Step 1

Simplify the right side.

Simplify $\log (4) + \log (x - 3)$ .

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

$\log (3 x) = \log (4 (x - 3))$

Apply the distributive property.

$\log (3 x) = \log (4 x + 4 \cdot - 3)$

Multiply $4$ by $- 3$ .

$\log (3 x) = \log (4 x - 12)$

Step 2

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

$3 x = 4 x - 12$

Step 3

Solve for $x$ .

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

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

$3 x - 4 x = - 12$

Subtract $4 x$ from $3 x$ .

$- x = - 12$

Divide each term in $- x = - 12$ by $- 1$ and simplify.

Divide each term in $- x = - 12$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 12}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 12}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 12}{- 1}$

Simplify the right side.

Divide $- 12$ by $- 1$ .

$x = 12$