Solve for x log base 6 of x-6+ log base 6 of x-5- log base 6 of x=1

$\log_{6} (x - 6) + \log_{6} (x - 5) - \log_{6} (x) = 1$

Step 1

Simplify the left side.

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

$\log_{6} ((x - 6) (x - 5)) - \log_{6} (x) = 1$

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

$\log_{6} (\frac{(x - 6) (x - 5)}{x}) = 1$

Step 2

Rewrite $\log_{6} (\frac{(x - 6) (x - 5)}{x}) = 1$ 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 $b^{y} = x$ .

$6^{1} = \frac{(x - 6) (x - 5)}{x}$

Step 3

Cross multiply to remove the fraction.

$(x - 6) (x - 5) = 6 (x)$

Step 4

Multiply $6$ by $x$ .

$(x - 6) (x - 5) = 6 x$

Step 5

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

Subtract $6 x$ from both sides of the equation.

$(x - 6) (x - 5) - 6 x = 0$

Simplify each term.

Expand $(x - 6) (x - 5)$ using the FOIL Method.

Apply the distributive property.

$x (x - 5) - 6 (x - 5) - 6 x = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 5 - 6 (x - 5) - 6 x = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 5 - 6 x - 6 \cdot - 5 - 6 x = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 5 - 6 x - 6 \cdot - 5 - 6 x = 0$

Move $- 5$ to the left of $x$ .

$x^{2} - 5 \cdot x - 6 x - 6 \cdot - 5 - 6 x = 0$

Multiply $- 6$ by $- 5$ .

$x^{2} - 5 x - 6 x + 30 - 6 x = 0$

Subtract $6 x$ from $- 5 x$ .

$x^{2} - 11 x + 30 - 6 x = 0$

Subtract $6 x$ from $- 11 x$ .

$x^{2} - 17 x + 30 = 0$

Step 6

Factor $x^{2} - 17 x + 30$ 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 $30$ and whose sum is $- 17$ .

$- 15, - 2$

Write the factored form using these integers.

$(x - 15) (x - 2) = 0$

Step 7

Simplify $(x - 15) (x - 2)$ .

Expand $(x - 15) (x - 2)$ using the FOIL Method.

Apply the distributive property.

$x (x - 2) - 15 (x - 2) = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 2 - 15 (x - 2) = 0$

Apply the distributive property.

$x \cdot x + x \cdot - 2 - 15 x - 15 \cdot - 2 = 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 2 - 15 x - 15 \cdot - 2 = 0$

Move $- 2$ to the left of $x$ .

$x^{2} - 2 \cdot x - 15 x - 15 \cdot - 2 = 0$

Multiply $- 15$ by $- 2$ .

$x^{2} - 2 x - 15 x + 30 = 0$

Subtract $15 x$ from $- 2 x$ .

$x^{2} - 17 x + 30 = 0$

Step 8

Factor $x^{2} - 17 x + 30$ 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 $30$ and whose sum is $- 17$ .

$- 15, - 2$

Write the factored form using these integers.

$(x - 15) (x - 2) = 0$

Step 9

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

$x - 15 = 0$

$x - 2 = 0$

Step 10

Set $x - 15$ equal to $0$ and solve for $x$ .

Set $x - 15$ equal to $0$ .

$x - 15 = 0$

Add $15$ to both sides of the equation.

$x = 15$

Step 11

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.

$x = 2$

Step 12

The final solution is all the values that make $(x - 15) (x - 2) = 0$ true.

$x = 15, 2$

Step 13

Exclude the solutions that do not make $\log_{6} (x - 6) + \log_{6} (x - 5) - \log_{6} (x) = 1$ true.

$x = 15$

Solve for x log base 6 of x-6+ log base 6 of x-5- log base 6 of x=1

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