Solve for x log base 2 of x^2+2- log base 2 of 3x+6=0

$\log_{2} (x^{2} + 2) - \log_{2} (3 x + 6) = 0$

Simplify the left side.

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{2} (\frac{x^{2} + 2}{3 x + 6}) = 0$

Factor $3$ out of $3 x + 6$ .

Factor $3$ out of $3 x$ .

$\log_{2} (\frac{x^{2} + 2}{3 (x) + 6}) = 0$

Factor $3$ out of $6$ .

$\log_{2} (\frac{x^{2} + 2}{3 x + 3 \cdot 2}) = 0$

Factor $3$ out of $3 x + 3 \cdot 2$ .

$\log_{2} (\frac{x^{2} + 2}{3 (x + 2)}) = 0$

Rewrite $\log_{2} (\frac{x^{2} + 2}{3 (x + 2)}) = 0$ 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 $b^{y} = x$ .

$2^{0} = \frac{x^{2} + 2}{3 (x + 2)}$

Cross multiply to remove the fraction.

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

Simplify $2^{0} (3 (x + 2))$ .

Simplify the expression.

Anything raised to $0$ is $1$ .

$x^{2} + 2 = 1 (3 (x + 2))$

Multiply $3 (x + 2)$ by $1$ .

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

Apply the distributive property.

$x^{2} + 2 = 3 x + 3 \cdot 2$

Multiply $3$ by $2$ .

$x^{2} + 2 = 3 x + 6$

Subtract $3 x$ from both sides of the equation.

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

Factor $x^{2} + 2 - 3 x$ 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 $2$ and whose sum is $- 3$ .

$- 2, - 1$

Write the factored form using these integers.

$(x - 2) (x - 1) = 6$

Simplify $(x - 2) (x - 1)$ .

Expand $(x - 2) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$x (x - 1) - 2 (x - 1) = 6$

Apply the distributive property.

$x \cdot x + x \cdot - 1 - 2 (x - 1) = 6$

Apply the distributive property.

$x \cdot x + x \cdot - 1 - 2 x - 2 \cdot - 1 = 6$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 - 2 x - 2 \cdot - 1 = 6$

Move $- 1$ to the left of $x$ .

$x^{2} - 1 \cdot x - 2 x - 2 \cdot - 1 = 6$

Rewrite $- 1 x$ as $- x$ .

$x^{2} - x - 2 x - 2 \cdot - 1 = 6$

Multiply $- 2$ by $- 1$ .

$x^{2} - x - 2 x + 2 = 6$

Subtract $2 x$ from $- x$ .

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

Subtract $6$ from both sides of the equation.

$x^{2} - 3 x + 2 - 6 = 0$

Subtract $6$ from $2$ .

$x^{2} - 3 x - 4 = 0$

Factor $x^{2} - 3 x - 4$ 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 $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$(x - 4) (x + 1) = 0$

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

$x - 4 = 0$

$x + 1 = 0$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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 $(x - 4) (x + 1) = 0$ true.

$x = 4, - 1$

Solve for x log base 2 of x^2+2- log base 2 of 3x+6=0

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