Solve for x log base -2x-3 of 3x^2+x-19=2

$\log_{- 2 x - 3} (3 x^{2} + x - 19) = 2$

Step 1

Rewrite $\log_{- 2 x - 3} (3 x^{2} + x - 19) = 2$ 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 $b^{y} = x$ .

${(- 2 x - 3)}^{2} = 3 x^{2} + x - 19$

Step 2

Solve for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$3 x^{2} + x - 19 = {(- 2 x - 3)}^{2}$

Simplify ${(- 2 x - 3)}^{2}$ .

Rewrite.

$3 x^{2} + x - 19 = 0 + 0 + {(- 2 x - 3)}^{2}$

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

$3 x^{2} + x - 19 = (- 2 x - 3) (- 2 x - 3)$

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

Apply the distributive property.

$3 x^{2} + x - 19 = - 2 x (- 2 x - 3) - 3 (- 2 x - 3)$

Apply the distributive property.

$3 x^{2} + x - 19 = - 2 x (- 2 x) - 2 x \cdot - 3 - 3 (- 2 x - 3)$

Apply the distributive property.

$3 x^{2} + x - 19 = - 2 x (- 2 x) - 2 x \cdot - 3 - 3 (- 2 x) - 3 \cdot - 3$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$3 x^{2} + x - 19 = - 2 \cdot - 2 x \cdot x - 2 x \cdot - 3 - 3 (- 2 x) - 3 \cdot - 3$

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

Move $x$ .

$3 x^{2} + x - 19 = - 2 \cdot - 2 (x \cdot x) - 2 x \cdot - 3 - 3 (- 2 x) - 3 \cdot - 3$

Multiply $x$ by $x$ .

$3 x^{2} + x - 19 = - 2 \cdot - 2 x^{2} - 2 x \cdot - 3 - 3 (- 2 x) - 3 \cdot - 3$

Multiply $- 2$ by $- 2$ .

$3 x^{2} + x - 19 = 4 x^{2} - 2 x \cdot - 3 - 3 (- 2 x) - 3 \cdot - 3$

Multiply $- 3$ by $- 2$ .

$3 x^{2} + x - 19 = 4 x^{2} + 6 x - 3 (- 2 x) - 3 \cdot - 3$

Multiply $- 2$ by $- 3$ .

$3 x^{2} + x - 19 = 4 x^{2} + 6 x + 6 x - 3 \cdot - 3$

Multiply $- 3$ by $- 3$ .

$3 x^{2} + x - 19 = 4 x^{2} + 6 x + 6 x + 9$

Add $6 x$ and $6 x$ .

$3 x^{2} + x - 19 = 4 x^{2} + 12 x + 9$

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

Subtract $4 x^{2}$ from both sides of the equation.

$3 x^{2} + x - 19 - 4 x^{2} = 12 x + 9$

Subtract $12 x$ from both sides of the equation.

$3 x^{2} + x - 19 - 4 x^{2} - 12 x = 9$

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

$- x^{2} + x - 19 - 12 x = 9$

Subtract $12 x$ from $x$ .

$- x^{2} - 11 x - 19 = 9$

Subtract $9$ from both sides of the equation.

$- x^{2} - 11 x - 19 - 9 = 0$

Subtract $9$ from $- 19$ .

$- x^{2} - 11 x - 28 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} - 11 x - 28$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) - 11 x - 28 = 0$

Factor $- 1$ out of $- 11 x$ .

$- (x^{2}) - (11 x) - 28 = 0$

Rewrite $- 28$ as $- 1 (28)$ .

$- (x^{2}) - (11 x) - 1 \cdot 28 = 0$

Factor $- 1$ out of $- (x^{2}) - (11 x)$ .

$- (x^{2} + 11 x) - 1 \cdot 28 = 0$

Factor $- 1$ out of $- (x^{2} + 11 x) - 1 (28)$ .

$- (x^{2} + 11 x + 28) = 0$

Factor.

Factor $x^{2} + 11 x + 28$ 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 $28$ and whose sum is $11$ .

$4, 7$

Write the factored form using these integers.

$- ((x + 4) (x + 7)) = 0$

Remove unnecessary parentheses.

$- (x + 4) (x + 7) = 0$

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

$x + 4 = 0$

$x + 7 = 0$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $- (x + 4) (x + 7) = 0$ true.

$x = - 4, - 7$

Solve for x log base -2x-3 of 3x^2+x-19=2

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