Solve for x log of 2x=1- log of x+4

$\log (2 x) = 1 - \log (x + 4)$

Move all the terms containing a logarithm to the left side of the equation.

$\log (2 x) + \log (x + 4) = 1$

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

$\log (2 x (x + 4)) = 1$

Apply the distributive property.

$\log (2 x \cdot x + 2 x \cdot 4) = 1$

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

Move $x$ .

$\log (2 (x \cdot x) + 2 x \cdot 4) = 1$

Multiply $x$ by $x$ .

$\log (2 x^{2} + 2 x \cdot 4) = 1$

Multiply $4$ by $2$ .

$\log (2 x^{2} + 8 x) = 1$

Rewrite $\log (2 x^{2} + 8 x) = 1$ 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 $b^{y} = x$ .

$10^{1} = 2 x^{2} + 8 x$

Solve for $x$ .

Rewrite the equation as $2 x^{2} + 8 x = 10^{1}$ .

$2 x^{2} + 8 x = 10$

Subtract $10$ from both sides of the equation.

$2 x^{2} + 8 x - 10 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} + 8 x - 10$ .

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

$2 (x^{2}) + 8 x - 10 = 0$

Factor $2$ out of $8 x$ .

$2 (x^{2}) + 2 (4 x) - 10 = 0$

Factor $2$ out of $- 10$ .

$2 x^{2} + 2 (4 x) + 2 \cdot - 5 = 0$

Factor $2$ out of $2 x^{2} + 2 (4 x)$ .

$2 (x^{2} + 4 x) + 2 \cdot - 5 = 0$

Factor $2$ out of $2 (x^{2} + 4 x) + 2 \cdot - 5$ .

$2 (x^{2} + 4 x - 5) = 0$

Factor.

Factor $x^{2} + 4 x - 5$ 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 $- 5$ and whose sum is $4$ .

$- 1, 5$

Write the factored form using these integers.

$2 ((x - 1) (x + 5)) = 0$

Remove unnecessary parentheses.

$2 (x - 1) (x + 5) = 0$

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

$x - 1 = 0$

$x + 5 = 0$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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.

$x = - 5$

The final solution is all the values that make $2 (x - 1) (x + 5) = 0$ true.

$x = 1, - 5$

Exclude the solutions that do not make $\log (2 x) = 1 - \log (x + 4)$ true.

$x = 1$

Solve for x log of 2x=1- log of x+4

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