Solve for x log base 6 of 2x- log base 6 of 3 = log base 6 of x-1

$\log_{6} (2 x) - \log_{6} (3) = \log_{6} (x - 1)$

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{6} (\frac{2 x}{3}) = \log_{6} (x - 1)$

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

$\frac{2 x}{3} = x - 1$

Solve for $x$ .

Multiply both sides by $3$ .

$\frac{2 x}{3} \cdot 3 = (x - 1) \cdot 3$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{2 x}{3} \cdot 3 = (x - 1) \cdot 3$

Rewrite the expression.

$2 x = (x - 1) \cdot 3$

Simplify the right side.

Simplify $(x - 1) \cdot 3$ .

Apply the distributive property.

$2 x = x \cdot 3 - 1 \cdot 3$

Simplify the expression.

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

$2 x = 3 \cdot x - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$2 x = 3 x - 3$

Solve for $x$ .

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

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

$2 x - 3 x = - 3$

Subtract $3 x$ from $2 x$ .

$- x = - 3$

Divide each term in $- x = - 3$ by $- 1$ and simplify.

Divide each term in $- x = - 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 3}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x = 3$