Evaluate log base 5 of 2x+1 = log base 5 of 4x-7

$\log_{5} (2 x + 1) = \log_{5} (4 x - 7)$

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

$2 x + 1 = 4 x - 7$

Solve for $x$ .

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

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

$2 x + 1 - 4 x = - 7$

Subtract $4 x$ from $2 x$ .

$- 2 x + 1 = - 7$

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

Subtract $1$ from both sides of the equation.

$- 2 x = - 7 - 1$

Subtract $1$ from $- 7$ .

$- 2 x = - 8$

Divide each term in $- 2 x = - 8$ by $- 2$ and simplify.

Divide each term in $- 2 x = - 8$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{- 8}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{- 8}{- 2}$

Divide $x$ by $1$ .

$x = \frac{- 8}{- 2}$

Simplify the right side.

Divide $- 8$ by $- 2$ .

$x = 4$