Solve for x log of x+9 = log of 2x-7

$\log (x + 9) = \log (2 x - 7)$

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

$x + 9 = 2 x - 7$

Solve for $x$ .

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

Subtract $2 x$ from both sides of the equation.

$x + 9 - 2 x = - 7$

Subtract $2 x$ from $x$ .

$- x + 9 = - 7$

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

Subtract $9$ from both sides of the equation.

$- x = - 7 - 9$

Subtract $9$ from $- 7$ .

$- x = - 16$

Divide each term in $- x = - 16$ by $- 1$ and simplify.

Divide each term in $- x = - 16$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 16}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 16}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 16}{- 1}$

Simplify the right side.

Divide $- 16$ by $- 1$ .

$x = 16$