Solve for x log of 4-x = log of x+8+ log of 2x+13

$\log (4 - x) = \log (x + 8) + \log (2 x + 13)$

Step 1

Simplify the right side.

Simplify $\log (x + 8) + \log (2 x + 13)$ .

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

$\log (4 - x) = \log ((x + 8) (2 x + 13))$

Expand $(x + 8) (2 x + 13)$ using the FOIL Method.

Apply the distributive property.

$\log (4 - x) = \log (x (2 x + 13) + 8 (2 x + 13))$

Apply the distributive property.

$\log (4 - x) = \log (x (2 x) + x \cdot 13 + 8 (2 x + 13))$

Apply the distributive property.

$\log (4 - x) = \log (x (2 x) + x \cdot 13 + 8 (2 x) + 8 \cdot 13)$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$\log (4 - x) = \log (2 x \cdot x + x \cdot 13 + 8 (2 x) + 8 \cdot 13)$

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

Move $x$ .

$\log (4 - x) = \log (2 (x \cdot x) + x \cdot 13 + 8 (2 x) + 8 \cdot 13)$

Multiply $x$ by $x$ .

$\log (4 - x) = \log (2 x^{2} + x \cdot 13 + 8 (2 x) + 8 \cdot 13)$

Move $13$ to the left of $x$ .

$\log (4 - x) = \log (2 x^{2} + 13 \cdot x + 8 (2 x) + 8 \cdot 13)$

Multiply $2$ by $8$ .

$\log (4 - x) = \log (2 x^{2} + 13 x + 16 x + 8 \cdot 13)$

Multiply $8$ by $13$ .

$\log (4 - x) = \log (2 x^{2} + 13 x + 16 x + 104)$

Add $13 x$ and $16 x$ .

$\log (4 - x) = \log (2 x^{2} + 29 x + 104)$

Step 2

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$4 - x = 2 x^{2} + 29 x + 104$

Step 3

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.

$2 x^{2} + 29 x + 104 = 4 - x$

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

Add $x$ to both sides of the equation.

$2 x^{2} + 29 x + 104 + x = 4$

Add $29 x$ and $x$ .

$2 x^{2} + 30 x + 104 = 4$

Subtract $4$ from both sides of the equation.

$2 x^{2} + 30 x + 104 - 4 = 0$

Subtract $4$ from $104$ .

$2 x^{2} + 30 x + 100 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 30 x + 100$ .

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

$2 (x^{2}) + 30 x + 100 = 0$

Factor $2$ out of $30 x$ .

$2 (x^{2}) + 2 (15 x) + 100 = 0$

Factor $2$ out of $100$ .

$2 x^{2} + 2 (15 x) + 2 \cdot 50 = 0$

Factor $2$ out of $2 x^{2} + 2 (15 x)$ .

$2 (x^{2} + 15 x) + 2 \cdot 50 = 0$

Factor $2$ out of $2 (x^{2} + 15 x) + 2 \cdot 50$ .

$2 (x^{2} + 15 x + 50) = 0$

Factor.

Factor $x^{2} + 15 x + 50$ 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 $50$ and whose sum is $15$ .

$5, 10$

Write the factored form using these integers.

$2 ((x + 5) (x + 10)) = 0$

Remove unnecessary parentheses.

$2 (x + 5) (x + 10) = 0$

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

$x + 5 = 0$

$x + 10 = 0$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Set $x + 10$ equal to $0$ and solve for $x$ .

Set $x + 10$ equal to $0$ .

$x + 10 = 0$

Subtract $10$ from both sides of the equation.

$x = - 10$

The final solution is all the values that make $2 (x + 5) (x + 10) = 0$ true.

$x = - 5, - 10$

Step 4

Exclude the solutions that do not make $\log (4 - x) = \log (x + 8) + \log (2 x + 13)$ true.

$x = - 5$

Solve for x log of 4-x = log of x+8+ log of 2x+13

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