Solve for a log base 7 of 6a+4 = log base 7 of 9a-5

$\log_{7} (6 a + 4) = \log_{7} (9 a - 5)$

Step 1

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

$6 a + 4 = 9 a - 5$

Step 2

Solve for $a$ .

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

Subtract $9 a$ from both sides of the equation.

$6 a + 4 - 9 a = - 5$

Subtract $9 a$ from $6 a$ .

$- 3 a + 4 = - 5$

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

Subtract $4$ from both sides of the equation.

$- 3 a = - 5 - 4$

Subtract $4$ from $- 5$ .

$- 3 a = - 9$

Divide each term in $- 3 a = - 9$ by $- 3$ and simplify.

Divide each term in $- 3 a = - 9$ by $- 3$ .

$\frac{- 3 a}{- 3} = \frac{- 9}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 a}{- 3} = \frac{- 9}{- 3}$

Divide $a$ by $1$ .

$a = \frac{- 9}{- 3}$

Simplify the right side.

Divide $- 9$ by $- 3$ .

$a = 3$