Solve for x 2 log base 3 of x+4- log base 3 of 9=2

$2 \log_{3} (x + 4) - \log_{3} (9) = 2$

Simplify each term.

Logarithm base $3$ of $9$ is $2$ .

$2 \log_{3} (x + 4) - 1 \cdot 2 = 2$

Multiply $- 1$ by $2$ .

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

Move all terms not containing $\log_{3} (x + 4)$ to the right side of the equation.

Add $2$ to both sides of the equation.

$2 \log_{3} (x + 4) = 2 + 2$

Add $2$ and $2$ .

$2 \log_{3} (x + 4) = 4$

Divide each term in $2 \log_{3} (x + 4) = 4$ by $2$ and simplify.

Divide each term in $2 \log_{3} (x + 4) = 4$ by $2$ .

$\frac{2 \log_{3} (x + 4)}{2} = \frac{4}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 \log_{3} (x + 4)}{2} = \frac{4}{2}$

Divide $\log_{3} (x + 4)$ by $1$ .

$\log_{3} (x + 4) = \frac{4}{2}$

Simplify the right side.

Divide $4$ by $2$ .

$\log_{3} (x + 4) = 2$

Rewrite $\log_{3} (x + 4) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$3^{2} = x + 4$

Solve for $x$ .

Rewrite the equation as $x + 4 = 3^{2}$ .

$x + 4 = 3^{2}$

Raise $3$ to the power of $2$ .

$x + 4 = 9$

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

Subtract $4$ from both sides of the equation.

$x = 9 - 4$

Subtract $4$ from $9$ .

$x = 5$

Solve for x 2 log base 3 of x+4- log base 3 of 9=2

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