Solve for w log base 2 of 2w^2- log base 2 of w+3=3

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

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

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

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

$2^{3} = \frac{2 w^{2}}{w + 3}$

Cross multiply to remove the fraction.

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

Simplify $2^{3} (w + 3)$ .

Raise $2$ to the power of $3$ .

$2 w^{2} = 8 (w + 3)$

Apply the distributive property.

$2 w^{2} = 8 w + 8 \cdot 3$

Multiply $8$ by $3$ .

$2 w^{2} = 8 w + 24$

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

$2 w^{2} - 8 w = 24$

Factor $2 w$ out of $2 w^{2} - 8 w$ .

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

$2 w (w) - 8 w = 24$

Factor $2 w$ out of $- 8 w$ .

$2 w (w) + 2 w (- 4) = 24$

Factor $2 w$ out of $2 w (w) + 2 w (- 4)$ .

$2 w (w - 4) = 24$

Divide each term in $2 w (w - 4) = 24$ by $2$ and simplify.

Divide each term in $2 w (w - 4) = 24$ by $2$ .

$\frac{2 w (w - 4)}{2} = \frac{24}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 w (w - 4)}{2} = \frac{24}{2}$

Divide $w (w - 4)$ by $1$ .

$w (w - 4) = \frac{24}{2}$

Apply the distributive property.

$w \cdot w + w \cdot - 4 = \frac{24}{2}$

Simplify the expression.

Multiply $w$ by $w$ .

$w^{2} + w \cdot - 4 = \frac{24}{2}$

Move $- 4$ to the left of $w$ .

$w^{2} - 4 w = \frac{24}{2}$

Simplify the right side.

Divide $24$ by $2$ .

$w^{2} - 4 w = 12$

Subtract $12$ from both sides of the equation.

$w^{2} - 4 w - 12 = 0$

Factor $w^{2} - 4 w - 12$ 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 $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Write the factored form using these integers.

$(w - 6) (w + 2) = 0$

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

$w - 6 = 0$

$w + 2 = 0$

Set $w - 6$ equal to $0$ and solve for $w$ .

Set $w - 6$ equal to $0$ .

$w - 6 = 0$

Add $6$ to both sides of the equation.

$w = 6$

Set $w + 2$ equal to $0$ and solve for $w$ .

Set $w + 2$ equal to $0$ .

$w + 2 = 0$

Subtract $2$ from both sides of the equation.

$w = - 2$

The final solution is all the values that make $(w - 6) (w + 2) = 0$ true.

$w = 6, - 2$

Solve for w log base 2 of 2w^2- log base 2 of w+3=3

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