Solve for u log base 9 of 3u+14- log base 9 of 5 = log base 9 of 2u

$\log_{9} (3 u + 14) - \log_{9} (5) = \log_{9} (2 u)$

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

$\log_{9} (\frac{3 u + 14}{5}) = \log_{9} (2 u)$

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

$\frac{3 u + 14}{5} = 2 u$

Solve for $u$ .

Multiply both sides by $5$ .

$\frac{3 u + 14}{5} \cdot 5 = 2 u \cdot 5$

Simplify.

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{3 u + 14}{5} \cdot 5 = 2 u \cdot 5$

Rewrite the expression.

$3 u + 14 = 2 u \cdot 5$

Simplify the right side.

Multiply $5$ by $2$ .

$3 u + 14 = 10 u$

Solve for $u$ .

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

Subtract $10 u$ from both sides of the equation.

$3 u + 14 - 10 u = 0$

Subtract $10 u$ from $3 u$ .

$- 7 u + 14 = 0$

Subtract $14$ from both sides of the equation.

$- 7 u = - 14$

Divide each term in $- 7 u = - 14$ by $- 7$ and simplify.

Divide each term in $- 7 u = - 14$ by $- 7$ .

$\frac{- 7 u}{- 7} = \frac{- 14}{- 7}$

Simplify the left side.

Cancel the common factor of $- 7$ .

Cancel the common factor.

$\frac{- 7 u}{- 7} = \frac{- 14}{- 7}$

Divide $u$ by $1$ .

$u = \frac{- 14}{- 7}$

Simplify the right side.

Divide $- 14$ by $- 7$ .

$u = 2$