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

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

Step 1

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

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

Step 2

Solve for $x$ .

Simplify the left side.

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

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

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

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

Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $- 3$ by $3$ .

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

Multiply $2$ by $- 3$ .

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

Add $- 9 x$ and $2 x$ .

$\log_{9} (3 x^{2} - 7 x - 6) = 0$

Rewrite $\log_{9} (3 x^{2} - 7 x - 6) = 0$ 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 $b^{y} = x$ .

$9^{0} = 3 x^{2} - 7 x - 6$

Solve for $x$ .

Rewrite the equation as $3 x^{2} - 7 x - 6 = 9^{0}$ .

$3 x^{2} - 7 x - 6 = 9^{0}$

Anything raised to $0$ is $1$ .

$3 x^{2} - 7 x - 6 = 1$

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

Subtract $1$ from both sides of the equation.

$3 x^{2} - 7 x - 6 - 1 = 0$

Subtract $1$ from $- 6$ .

$3 x^{2} - 7 x - 7 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 3$ , $b = - 7$ , and $c = - 7$ into the quadratic formula and solve for $x$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (3 \cdot - 7)}}{2 \cdot 3}$

Simplify.

Simplify the numerator.

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 3 \cdot - 7}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 7$ .

Multiply $- 4$ by $3$ .

$x = \frac{7 \pm \sqrt{49 - 12 \cdot - 7}}{2 \cdot 3}$

Multiply $- 12$ by $- 7$ .

$x = \frac{7 \pm \sqrt{49 + 84}}{2 \cdot 3}$

Add $49$ and $84$ .

$x = \frac{7 \pm \sqrt{133}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{7 \pm \sqrt{133}}{6}$

The final answer is the combination of both solutions.

$x = \frac{7 + \sqrt{133}}{6}, \frac{7 - \sqrt{133}}{6}$

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

$x = \frac{7 + \sqrt{133}}{6}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = \frac{7 + \sqrt{133}}{6}$

Decimal Form:

$x = 3.08876043 \dots$

Step 4

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

Related Questions

Support LCMGCF.com by disabling your adblocker.