Solve for x log base 6 of 3x-11+ log base 6 of 2 = log base 6 of 4x-8

$\log_{6} (3 x - 11) + \log_{6} (2) = \log_{6} (4 x - 8)$

Simplify the left side.

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

$\log_{6} ((3 x - 11) \cdot 2) = \log_{6} (4 x - 8)$

Apply the distributive property.

$\log_{6} (3 x \cdot 2 - 11 \cdot 2) = \log_{6} (4 x - 8)$

Multiply.

Multiply $2$ by $3$ .

$\log_{6} (6 x - 11 \cdot 2) = \log_{6} (4 x - 8)$

Multiply $- 11$ by $2$ .

$\log_{6} (6 x - 22) = \log_{6} (4 x - 8)$

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

$6 x - 22 = 4 x - 8$

Solve for $x$ .

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

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

$6 x - 22 - 4 x = - 8$

Subtract $4 x$ from $6 x$ .

$2 x - 22 = - 8$

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

Add $22$ to both sides of the equation.

$2 x = - 8 + 22$

Add $- 8$ and $22$ .

$2 x = 14$

Divide each term in $2 x = 14$ by $2$ and simplify.

Divide each term in $2 x = 14$ by $2$ .

$\frac{2 x}{2} = \frac{14}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{14}{2}$

Divide $x$ by $1$ .

$x = \frac{14}{2}$

Simplify the right side.

Divide $14$ by $2$ .

$x = 7$