Solve for x log base 5 of x = log base 5 of 8- log base 5 of 6-x

$\log_{5} (x) = \log_{5} (8) - \log_{5} (6 - x)$

Simplify the right side.

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

$\log_{5} (x) = \log_{5} (\frac{8}{6 - x})$

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

$x = \frac{8}{6 - x}$

Solve for $x$ .

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.

$1, 6 - x$

Remove parentheses.

$1, 6 - x$

The LCM of one and any expression is the expression.

$6 - x$

Multiply each term in $x = \frac{8}{6 - x}$ by $6 - x$ to eliminate the fractions.

Multiply each term in $x = \frac{8}{6 - x}$ by $6 - x$ .

$x (6 - x) = \frac{8}{6 - x} (6 - x)$

Simplify the left side.

Simplify by multiplying through.

Apply the distributive property.

$x \cdot 6 + x (- x) = \frac{8}{6 - x} (6 - x)$

Reorder.

Move $6$ to the left of $x$ .

$6 \cdot x + x (- x) = \frac{8}{6 - x} (6 - x)$

Rewrite using the commutative property of multiplication.

$6 \cdot x - x \cdot x = \frac{8}{6 - x} (6 - x)$

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

Move $x$ .

$6 x - (x \cdot x) = \frac{8}{6 - x} (6 - x)$

Multiply $x$ by $x$ .

$6 x - x^{2} = \frac{8}{6 - x} (6 - x)$

Simplify the right side.

Cancel the common factor of $6 - x$ .

Cancel the common factor.

$6 x - x^{2} = \frac{8}{6 - x} (6 - x)$

Rewrite the expression.

$6 x - x^{2} = 8$

Solve the equation.

Subtract $8$ from both sides of the equation.

$6 x - x^{2} - 8 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $6 x - x^{2} - 8$ .

Reorder $6 x$ and $- x^{2}$ .

$- x^{2} + 6 x - 8 = 0$

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

$- (x^{2}) + 6 x - 8 = 0$

Factor $- 1$ out of $6 x$ .

$- (x^{2}) - (- 6 x) - 8 = 0$

Rewrite $- 8$ as $- 1 (8)$ .

$- (x^{2}) - (- 6 x) - 1 \cdot 8 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 6 x)$ .

$- (x^{2} - 6 x) - 1 \cdot 8 = 0$

Factor $- 1$ out of $- (x^{2} - 6 x) - 1 (8)$ .

$- (x^{2} - 6 x + 8) = 0$

Factor.

Factor $x^{2} - 6 x + 8$ 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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Write the factored form using these integers.

$- ((x - 4) (x - 2)) = 0$

Remove unnecessary parentheses.

$- (x - 4) (x - 2) = 0$

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

$x - 4 = 0$

$x - 2 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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.

$x = 2$

The final solution is all the values that make $- (x - 4) (x - 2) = 0$ true.

$x = 4, 2$

Solve for x log base 5 of x = log base 5 of 8- log base 5 of 6-x

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