Solve for x log base 2x-3 of 3x^2-2x-12=2

$\log_{2 x - 3} (3 x^{2} - 2 x - 12) = 2$

Step 1

Rewrite $\log_{2 x - 3} (3 x^{2} - 2 x - 12) = 2$ 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 $b^{y} = x$ .

${(2 x - 3)}^{2} = 3 x^{2} - 2 x - 12$

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.

$3 x^{2} - 2 x - 12 = {(2 x - 3)}^{2}$

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

Rewrite.

$3 x^{2} - 2 x - 12 = 0 + 0 + {(2 x - 3)}^{2}$

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

$3 x^{2} - 2 x - 12 = (2 x - 3) (2 x - 3)$

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

Apply the distributive property.

$3 x^{2} - 2 x - 12 = 2 x (2 x - 3) - 3 (2 x - 3)$

Apply the distributive property.

$3 x^{2} - 2 x - 12 = 2 x (2 x) + 2 x \cdot - 3 - 3 (2 x - 3)$

Apply the distributive property.

$3 x^{2} - 2 x - 12 = 2 x (2 x) + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$3 x^{2} - 2 x - 12 = 2 \cdot 2 x \cdot x + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3$

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

Move $x$ .

$3 x^{2} - 2 x - 12 = 2 \cdot 2 (x \cdot x) + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3$

Multiply $x$ by $x$ .

$3 x^{2} - 2 x - 12 = 2 \cdot 2 x^{2} + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3$

Multiply $2$ by $2$ .

$3 x^{2} - 2 x - 12 = 4 x^{2} + 2 x \cdot - 3 - 3 (2 x) - 3 \cdot - 3$

Multiply $- 3$ by $2$ .

$3 x^{2} - 2 x - 12 = 4 x^{2} - 6 x - 3 (2 x) - 3 \cdot - 3$

Multiply $2$ by $- 3$ .

$3 x^{2} - 2 x - 12 = 4 x^{2} - 6 x - 6 x - 3 \cdot - 3$

Multiply $- 3$ by $- 3$ .

$3 x^{2} - 2 x - 12 = 4 x^{2} - 6 x - 6 x + 9$

Subtract $6 x$ from $- 6 x$ .

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

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

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

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

Add $12 x$ to both sides of the equation.

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

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

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

Add $- 2 x$ and $12 x$ .

$- x^{2} + 10 x - 12 = 9$

Subtract $9$ from both sides of the equation.

$- x^{2} + 10 x - 12 - 9 = 0$

Subtract $9$ from $- 12$ .

$- x^{2} + 10 x - 21 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 10 x - 21$ .

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

$- (x^{2}) + 10 x - 21 = 0$

Factor $- 1$ out of $10 x$ .

$- (x^{2}) - (- 10 x) - 21 = 0$

Rewrite $- 21$ as $- 1 (21)$ .

$- (x^{2}) - (- 10 x) - 1 \cdot 21 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 10 x)$ .

$- (x^{2} - 10 x) - 1 \cdot 21 = 0$

Factor $- 1$ out of $- (x^{2} - 10 x) - 1 (21)$ .

$- (x^{2} - 10 x + 21) = 0$

Factor.

Factor $x^{2} - 10 x + 21$ 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 $21$ and whose sum is $- 10$ .

$- 7, - 3$

Write the factored form using these integers.

$- ((x - 7) (x - 3)) = 0$

Remove unnecessary parentheses.

$- (x - 7) (x - 3) = 0$

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

$x - 7 = 0$

$x - 3 = 0$

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

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $- (x - 7) (x - 3) = 0$ true.

$x = 7, 3$

Solve for x log base 2x-3 of 3x^2-2x-12=2

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