Solve for x 2 natural log of x- natural log of 3-x = natural log of 1/2+ natural log of 8

$2 \ln (x) - \ln (3 - x) = \ln (\frac{1}{2}) + \ln (8)$

Reorder $3$ and $- x$ .

$2 \ln (x) - \ln (- x + 3) = \ln (\frac{1}{2}) + \ln (8)$

Simplify the right side.

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

$2 \ln (x) - \ln (- x + 3) = \ln (\frac{1}{2} \cdot 8)$

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

$2 \ln (x) - \ln (- x + 3) = \ln (\frac{1}{2} \cdot (2 (4)))$

Cancel the common factor.

$2 \ln (x) - \ln (- x + 3) = \ln (\frac{1}{2} \cdot (2 \cdot 4))$

Rewrite the expression.

$2 \ln (x) - \ln (- x + 3) = \ln (4)$

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

$2 \ln (x) - \ln (- x + 3) - \ln (4) = 0$

Simplify the left side.

Simplify $2 \ln (x) - \ln (- x + 3) - \ln (4)$ .

Simplify $2 \ln (x)$ by moving $2$ inside the logarithm.

$\ln (x^{2}) - \ln (- x + 3) - \ln (4) = 0$

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

$\ln (\frac{x^{2}}{- x + 3}) - \ln (4) = 0$

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

$\ln (\frac{\frac{x^{2}}{- x + 3}}{4}) = 0$

Multiply the numerator by the reciprocal of the denominator.

$\ln (\frac{x^{2}}{- x + 3} \cdot \frac{1}{4}) = 0$

Multiply $\frac{x^{2}}{- x + 3}$ by $\frac{1}{4}$ .

$\ln (\frac{x^{2}}{(- x + 3) \cdot 4}) = 0$

Move $4$ to the left of $- x + 3$ .

$\ln (\frac{x^{2}}{4 (- x + 3)}) = 0$

Rewrite $\ln (\frac{x^{2}}{4 (- x + 3)}) = 0$ 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 $b^{y} = x$ .

$e^{0} = \frac{x^{2}}{4 (- x + 3)}$

Cross multiply to remove the fraction.

$x^{2} = e^{0} (4 (- x + 3))$

Simplify $e^{0} (4 (- x + 3))$ .

Simplify the expression.

Anything raised to $0$ is $1$ .

$x^{2} = 1 (4 (- x + 3))$

Multiply $4 (- x + 3)$ by $1$ .

$x^{2} = 4 (- x + 3)$

Apply the distributive property.

$x^{2} = 4 (- x) + 4 \cdot 3$

Multiply.

Multiply $- 1$ by $4$ .

$x^{2} = - 4 x + 4 \cdot 3$

Multiply $4$ by $3$ .

$x^{2} = - 4 x + 12$

Add $4 x$ to both sides of the equation.

$x^{2} + 4 x = 12$

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

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

$x \cdot x + 4 x = 12$

Factor $x$ out of $4 x$ .

$x \cdot x + x \cdot 4 = 12$

Factor $x$ out of $x \cdot x + x \cdot 4$ .

$x (x + 4) = 12$

Simplify $x (x + 4)$ .

Apply the distributive property.

$x \cdot x + x \cdot 4 = 12$

Simplify the expression.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 4 = 12$

Move $4$ to the left of $x$ .

$x^{2} + 4 x = 12$

Subtract $12$ from both sides of the equation.

$x^{2} + 4 x - 12 = 0$

Factor $x^{2} + 4 x - 12$ 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 $4$ .

$- 2, 6$

Write the factored form using these integers.

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

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

$x - 2 = 0$

$x + 6 = 0$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Set $x + 6$ equal to $0$ and solve for $x$ .

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Subtract $6$ from both sides of the equation.

$x = - 6$

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

$x = 2, - 6$

Exclude the solutions that do not make $2 \ln (x) - \ln (3 - x) = \ln (\frac{1}{2}) + \ln (8)$ true.

$x = 2$

Solve for x 2 natural log of x- natural log of 3-x = natural log of 1/2+ natural log of 8

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