Solve for x log base x+4 of 3x^2-6x=2

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

Step 1

Rewrite $\log_{x + 4} (3 x^{2} - 6 x) = 2$ 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 $b^{y} = x$ .

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

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.

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

Simplify ${(x + 4)}^{2}$ .

Rewrite ${(x + 4)}^{2}$ as $(x + 4) (x + 4)$ .

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

Expand $(x + 4) (x + 4)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Move $4$ to the left of $x$ .

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

Multiply $4$ by $4$ .

$3 x^{2} - 6 x = x^{2} + 4 x + 4 x + 16$

Add $4 x$ and $4 x$ .

$3 x^{2} - 6 x = x^{2} + 8 x + 16$

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

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

$3 x^{2} - 6 x - x^{2} = 8 x + 16$

Subtract $8 x$ from both sides of the equation.

$3 x^{2} - 6 x - x^{2} - 8 x = 16$

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

$2 x^{2} - 6 x - 8 x = 16$

Subtract $8 x$ from $- 6 x$ .

$2 x^{2} - 14 x = 16$

Subtract $16$ from both sides of the equation.

$2 x^{2} - 14 x - 16 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 14 x - 16$ .

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

$2 (x^{2}) - 14 x - 16 = 0$

Factor $2$ out of $- 14 x$ .

$2 (x^{2}) + 2 (- 7 x) - 16 = 0$

Factor $2$ out of $- 16$ .

$2 x^{2} + 2 (- 7 x) + 2 \cdot - 8 = 0$

Factor $2$ out of $2 x^{2} + 2 (- 7 x)$ .

$2 (x^{2} - 7 x) + 2 \cdot - 8 = 0$

Factor $2$ out of $2 (x^{2} - 7 x) + 2 \cdot - 8$ .

$2 (x^{2} - 7 x - 8) = 0$

Factor.

Factor $x^{2} - 7 x - 8$ 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 $- 8$ and whose sum is $- 7$ .

$- 8, 1$

Write the factored form using these integers.

$2 ((x - 8) (x + 1)) = 0$

Remove unnecessary parentheses.

$2 (x - 8) (x + 1) = 0$

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

$x - 8 = 0$

$x + 1 = 0$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $2 (x - 8) (x + 1) = 0$ true.

$x = 8, - 1$

Solve for x log base x+4 of 3x^2-6x=2

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