Find the Roots (Zeros) f(x) = log base 9 of 2x+3+ log base 9 of x-2

$f (x) = \log_{9} (2 x + 3) + \log_{9} (x - 2)$

Set $\log_{9} (2 x + 3) + \log_{9} (x - 2)$ equal to $0$ .

$\log_{9} (2 x + 3) + \log_{9} (x - 2) = 0$

Solve for $x$ .

Simplify the left side.

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

$\log_{9} ((2 x + 3) (x - 2)) = 0$

Expand $(2 x + 3) (x - 2)$ using the FOIL Method.

Apply the distributive property.

$\log_{9} (2 x (x - 2) + 3 (x - 2)) = 0$

Apply the distributive property.

$\log_{9} (2 x \cdot x + 2 x \cdot - 2 + 3 (x - 2)) = 0$

Apply the distributive property.

$\log_{9} (2 x \cdot x + 2 x \cdot - 2 + 3 x + 3 \cdot - 2) = 0$

Simplify and combine like terms.

Simplify each term.

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

Move $x$ .

$\log_{9} (2 (x \cdot x) + 2 x \cdot - 2 + 3 x + 3 \cdot - 2) = 0$

Multiply $x$ by $x$ .

$\log_{9} (2 x^{2} + 2 x \cdot - 2 + 3 x + 3 \cdot - 2) = 0$

Multiply $- 2$ by $2$ .

$\log_{9} (2 x^{2} - 4 x + 3 x + 3 \cdot - 2) = 0$

Multiply $3$ by $- 2$ .

$\log_{9} (2 x^{2} - 4 x + 3 x - 6) = 0$

Add $- 4 x$ and $3 x$ .

$\log_{9} (2 x^{2} - x - 6) = 0$

Rewrite $\log_{9} (2 x^{2} - x - 6) = 0$ 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 $b^{y} = x$ .

$9^{0} = 2 x^{2} - x - 6$

Solve for $x$ .

Rewrite the equation as $2 x^{2} - x - 6 = 9^{0}$ .

$2 x^{2} - x - 6 = 9^{0}$

Anything raised to $0$ is $1$ .

$2 x^{2} - x - 6 = 1$

Move all terms to the left side of the equation and simplify.

Subtract $1$ from both sides of the equation.

$2 x^{2} - x - 6 - 1 = 0$

Subtract $1$ from $- 6$ .

$2 x^{2} - x - 7 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 2$ , $b = - 1$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (2 \cdot - 7)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 2 \cdot - 7}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8 \cdot - 7}}{2 \cdot 2}$

Multiply $- 8$ by $- 7$ .

$x = \frac{1 \pm \sqrt{1 + 56}}{2 \cdot 2}$

Add $1$ and $56$ .

$x = \frac{1 \pm \sqrt{57}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{1 \pm \sqrt{57}}{4}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{57}}{4}, \frac{1 - \sqrt{57}}{4}$

Exclude the solutions that do not make $\log_{9} (2 x + 3) + \log_{9} (x - 2) = 0$ true.

$x = \frac{1 + \sqrt{57}}{4}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 + \sqrt{57}}{4}$

Decimal Form:

$x = 2.13745860 \dots$

Find the Roots (Zeros) f(x) = log base 9 of 2x+3+ log base 9 of x-2

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