Find All Complex Solutions log base 2 of x+6+ log base 2 of x+5=1

$\log_{2} (x + 6) + \log_{2} (x + 5) = 1$

Simplify the left side.

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{2} ((x + 6) (x + 5)) = 1$

Expand $(x + 6) (x + 5)$ using the FOIL Method.

Apply the distributive property.

$\log_{2} (x (x + 5) + 6 (x + 5)) = 1$

Apply the distributive property.

$\log_{2} (x \cdot x + x \cdot 5 + 6 (x + 5)) = 1$

Apply the distributive property.

$\log_{2} (x \cdot x + x \cdot 5 + 6 x + 6 \cdot 5) = 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\log_{2} (x^{2} + x \cdot 5 + 6 x + 6 \cdot 5) = 1$

Move $5$ to the left of $x$ .

$\log_{2} (x^{2} + 5 \cdot x + 6 x + 6 \cdot 5) = 1$

Multiply $6$ by $5$ .

$\log_{2} (x^{2} + 5 x + 6 x + 30) = 1$

Add $5 x$ and $6 x$ .

$\log_{2} (x^{2} + 11 x + 30) = 1$

Rewrite $\log_{2} (x^{2} + 11 x + 30) = 1$ 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 $b^{y} = x$ .

$2^{1} = x^{2} + 11 x + 30$

Solve for $x$ .

Rewrite the equation as $x^{2} + 11 x + 30 = 2^{1}$ .

$x^{2} + 11 x + 30 = 2$

Subtract $2$ from both sides of the equation.

$x^{2} + 11 x + 30 - 2 = 0$

Subtract $2$ from $30$ .

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

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.

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

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.

$x = - 4, - 7$

Exclude the solutions that do not make $\log_{2} (x + 6) + \log_{2} (x + 5) = 1$ true.

$x = - 4$

Find All Complex Solutions log base 2 of x+6+ log base 2 of x+5=1

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