Solve for x log base 9 of x-5+ log base 9 of x+3=1

$\log_{9} (x - 5) + \log_{9} (x + 3) = 1$

Simplify the left side.

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

$\log_{9} ((x - 5) (x + 3)) = 1$

Expand $(x - 5) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$\log_{9} (x (x + 3) - 5 (x + 3)) = 1$

Apply the distributive property.

$\log_{9} (x \cdot x + x \cdot 3 - 5 (x + 3)) = 1$

Apply the distributive property.

$\log_{9} (x \cdot x + x \cdot 3 - 5 x - 5 \cdot 3) = 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\log_{9} (x^{2} + x \cdot 3 - 5 x - 5 \cdot 3) = 1$

Move $3$ to the left of $x$ .

$\log_{9} (x^{2} + 3 \cdot x - 5 x - 5 \cdot 3) = 1$

Multiply $- 5$ by $3$ .

$\log_{9} (x^{2} + 3 x - 5 x - 15) = 1$

Subtract $5 x$ from $3 x$ .

$\log_{9} (x^{2} - 2 x - 15) = 1$

Rewrite $\log_{9} (x^{2} - 2 x - 15) = 1$ 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 $b^{y} = x$ .

$9^{1} = x^{2} - 2 x - 15$

Solve for $x$ .

Rewrite the equation as $x^{2} - 2 x - 15 = 9^{1}$ .

$x^{2} - 2 x - 15 = 9$

Subtract $9$ from both sides of the equation.

$x^{2} - 2 x - 15 - 9 = 0$

Subtract $9$ from $- 15$ .

$x^{2} - 2 x - 24 = 0$

Factor $x^{2} - 2 x - 24$ 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 $- 24$ and whose sum is $- 2$ .

$- 6, 4$

Write the factored form using these integers.

$(x - 6) (x + 4) = 0$

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

$x - 6 = 0$

$x + 4 = 0$

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

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

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(x - 6) (x + 4) = 0$ true.

$x = 6, - 4$

Exclude the solutions that do not make $\log_{9} (x - 5) + \log_{9} (x + 3) = 1$ true.

$x = 6$

Solve for x log base 9 of x-5+ log base 9 of x+3=1

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