Solve for x log base -2x-3 of 3x^2-18=2

$\log_{- 2 x - 3} (3 x^{2} - 18) = 2$

Step 1

Rewrite $\log_{- 2 x - 3} (3 x^{2} - 18) = 2$ 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 $b^{y} = x$ .

${(- 2 x - 3)}^{2} = 3 x^{2} - 18$

Step 2

Solve for $x$ .

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

Subtract $3 x^{2}$ from both sides of the equation.

${(- 2 x - 3)}^{2} - 3 x^{2} = - 18$

Simplify each term.

Rewrite ${(- 2 x - 3)}^{2}$ as $(- 2 x - 3) (- 2 x - 3)$ .

$(- 2 x - 3) (- 2 x - 3) - 3 x^{2} = - 18$

Expand $(- 2 x - 3) (- 2 x - 3)$ using the FOIL Method.

Apply the distributive property.

$- 2 x (- 2 x - 3) - 3 (- 2 x - 3) - 3 x^{2} = - 18$

Apply the distributive property.

$- 2 x (- 2 x) - 2 x \cdot - 3 - 3 (- 2 x - 3) - 3 x^{2} = - 18$

Apply the distributive property.

$- 2 x (- 2 x) - 2 x \cdot - 3 - 3 (- 2 x) - 3 \cdot - 3 - 3 x^{2} = - 18$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$- 2 \cdot - 2 x \cdot x - 2 x \cdot - 3 - 3 (- 2 x) - 3 \cdot - 3 - 3 x^{2} = - 18$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$- 2 \cdot - 2 (x \cdot x) - 2 x \cdot - 3 - 3 (- 2 x) - 3 \cdot - 3 - 3 x^{2} = - 18$

Multiply $x$ by $x$ .

$- 2 \cdot - 2 x^{2} - 2 x \cdot - 3 - 3 (- 2 x) - 3 \cdot - 3 - 3 x^{2} = - 18$

Multiply $- 2$ by $- 2$ .

$4 x^{2} - 2 x \cdot - 3 - 3 (- 2 x) - 3 \cdot - 3 - 3 x^{2} = - 18$

Multiply $- 3$ by $- 2$ .

$4 x^{2} + 6 x - 3 (- 2 x) - 3 \cdot - 3 - 3 x^{2} = - 18$

Multiply $- 2$ by $- 3$ .

$4 x^{2} + 6 x + 6 x - 3 \cdot - 3 - 3 x^{2} = - 18$

Multiply $- 3$ by $- 3$ .

$4 x^{2} + 6 x + 6 x + 9 - 3 x^{2} = - 18$

Add $6 x$ and $6 x$ .

$4 x^{2} + 12 x + 9 - 3 x^{2} = - 18$

Subtract $3 x^{2}$ from $4 x^{2}$ .

$x^{2} + 12 x + 9 = - 18$

Add $18$ to both sides of the equation.

$x^{2} + 12 x + 9 + 18 = 0$

Add $9$ and $18$ .

$x^{2} + 12 x + 27 = 0$

Factor $x^{2} + 12 x + 27$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $27$ and whose sum is $12$ .

$3, 9$

Write the factored form using these integers.

$(x + 3) (x + 9) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 3 = 0$

$x + 9 = 0$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set $x + 9$ equal to $0$ and solve for $x$ .

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Subtract $9$ from both sides of the equation.

$x = - 9$

The final solution is all the values that make $(x + 3) (x + 9) = 0$ true.

$x = - 3, - 9$

Solve for x log base -2x-3 of 3x^2-18=2

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