Solve for x log base 3 of -2x-3=2

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

Step 1

Rewrite $\log_{3} (- 2 x - 3) = 2$ 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 $b^{y} = x$ .

$3^{2} = - 2 x - 3$

Step 2

Solve for $x$ .

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

$- 2 x - 3 = 3^{2}$

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

$- 2 x - 3 = 9$

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

Add $3$ to both sides of the equation.

$- 2 x = 9 + 3$

Add $9$ and $3$ .

$- 2 x = 12$

Divide each term in $- 2 x = 12$ by $- 2$ and simplify.

Divide each term in $- 2 x = 12$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{12}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{12}{- 2}$

Divide $x$ by $1$ .

$x = \frac{12}{- 2}$

Simplify the right side.

Divide $12$ by $- 2$ .

$x = - 6$

Solve for x log base 3 of -2x-3=2

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