Solve for x natural log of x-6=2 natural log of 2- natural log of 10-x

$\ln (x - 6) = 2 \ln (2) - \ln (10 - x)$

Step 1

Simplify the right side.

Simplify $2 \ln (2) - \ln (10 - x)$ .

Simplify each term.

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

$\ln (x - 6) = \ln (2^{2}) - \ln (10 - x)$

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

$\ln (x - 6) = \ln (4) - \ln (10 - x)$

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

$\ln (x - 6) = \ln (\frac{4}{10 - x})$

Step 2

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

$x - 6 = \frac{4}{10 - x}$

Step 3

Solve for $x$ .

Add $6$ to both sides of the equation.

$x = \frac{4}{10 - x} + 6$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, 10 - x, 1$

Remove parentheses.

$1, 10 - x, 1$

The LCM of one and any expression is the expression.

$10 - x$

Multiply each term in $x = \frac{4}{10 - x} + 6$ by $10 - x$ to eliminate the fractions.

Multiply each term in $x = \frac{4}{10 - x} + 6$ by $10 - x$ .

$x (10 - x) = \frac{4}{10 - x} (10 - x) + 6 (10 - x)$

Simplify the left side.

Simplify by multiplying through.

Apply the distributive property.

$x \cdot 10 + x (- x) = \frac{4}{10 - x} (10 - x) + 6 (10 - x)$

Reorder.

Move $10$ to the left of $x$ .

$10 \cdot x + x (- x) = \frac{4}{10 - x} (10 - x) + 6 (10 - x)$

Rewrite using the commutative property of multiplication.

$10 \cdot x - x \cdot x = \frac{4}{10 - x} (10 - x) + 6 (10 - x)$

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

Move $x$ .

$10 x - (x \cdot x) = \frac{4}{10 - x} (10 - x) + 6 (10 - x)$

Multiply $x$ by $x$ .

$10 x - x^{2} = \frac{4}{10 - x} (10 - x) + 6 (10 - x)$

Simplify the right side.

Simplify each term.

Cancel the common factor of $10 - x$ .

Cancel the common factor.

$10 x - x^{2} = \frac{4}{10 - x} (10 - x) + 6 (10 - x)$

Rewrite the expression.

$10 x - x^{2} = 4 + 6 (10 - x)$

Apply the distributive property.

$10 x - x^{2} = 4 + 6 \cdot 10 + 6 (- x)$

Multiply $6$ by $10$ .

$10 x - x^{2} = 4 + 60 + 6 (- x)$

Multiply $- 1$ by $6$ .

$10 x - x^{2} = 4 + 60 - 6 x$

Add $4$ and $60$ .

$10 x - x^{2} = 64 - 6 x$

Solve the equation.

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

Add $6 x$ to both sides of the equation.

$10 x - x^{2} + 6 x = 64$

Add $10 x$ and $6 x$ .

$- x^{2} + 16 x = 64$

Subtract $64$ from both sides of the equation.

$- x^{2} + 16 x - 64 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 16 x - 64$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 16 x - 64 = 0$

Factor $- 1$ out of $16 x$ .

$- (x^{2}) - (- 16 x) - 64 = 0$

Rewrite $- 64$ as $- 1 (64)$ .

$- (x^{2}) - (- 16 x) - 1 \cdot 64 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 16 x)$ .

$- (x^{2} - 16 x) - 1 \cdot 64 = 0$

Factor $- 1$ out of $- (x^{2} - 16 x) - 1 (64)$ .

$- (x^{2} - 16 x + 64) = 0$

Factor using the perfect square rule.

Rewrite $64$ as $8^{2}$ .

$- (x^{2} - 16 x + 8^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$16 x = 2 \cdot x \cdot 8$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 8$ .

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

Divide each term in $- {(x - 8)}^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- {(x - 8)}^{2} = 0$ by $- 1$ .

$\frac{- {(x - 8)}^{2}}{- 1} = \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{{(x - 8)}^{2}}{1} = \frac{0}{- 1}$

Divide ${(x - 8)}^{2}$ by $1$ .

${(x - 8)}^{2} = \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

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

Set the $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Solve for x natural log of x-6=2 natural log of 2- natural log of 10-x

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