Solve for x log base 9 of x-2+ log base 9 of x=1/2

$\log_{9} (x - 2) + \log_{9} (x) = \frac{1}{2}$

Simplify the left side.

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

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

Apply the distributive property.

$\log_{9} (x \cdot x - 2 x) = \frac{1}{2}$

Multiply $x$ by $x$ .

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

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

$9^{\frac{1}{2}} = x^{2} - 2 x$

Solve for $x$ .

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

$x^{2} - 2 x = 9^{\frac{1}{2}}$

Simplify $9^{\frac{1}{2}}$ .

Simplify the expression.

Rewrite $9$ as $3^{2}$ .

$x^{2} - 2 x = {(3^{2})}^{\frac{1}{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{2} - 2 x = 3^{2 (\frac{1}{2})}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{2} - 2 x = 3^{2 (\frac{1}{2})}$

Rewrite the expression.

$x^{2} - 2 x = 3^{1}$

Evaluate the exponent.

$x^{2} - 2 x = 3$

Subtract $3$ from both sides of the equation.

$x^{2} - 2 x - 3 = 0$

Factor $x^{2} - 2 x - 3$ 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 $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$(x - 3) (x + 1) = 0$

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

$x - 3 = 0$

$x + 1 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 3) (x + 1) = 0$ true.

$x = 3, - 1$

Exclude the solutions that do not make $\log_{9} (x - 2) + \log_{9} (x) = \frac{1}{2}$ true.

$x = 3$

Solve for x log base 9 of x-2+ log base 9 of x=1/2

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