Find the Roots (Zeros) f(x)=(4x-13) natural log of (2x-7)(x-3)

$f (x) = (4 x - 13) \ln ((2 x - 7) (x - 3))$

Set $(4 x - 13) \ln ((2 x - 7) (x - 3))$ equal to $0$ .

$(4 x - 13) \ln ((2 x - 7) (x - 3)) = 0$

Solve for $x$ .

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

$4 x - 13 = 0$

$\ln ((2 x - 7) (x - 3)) = 0$

Set $4 x - 13$ equal to $0$ and solve for $x$ .

Set $4 x - 13$ equal to $0$ .

$4 x - 13 = 0$

Solve $4 x - 13 = 0$ for $x$ .

Add $13$ to both sides of the equation.

$4 x = 13$

Divide each term in $4 x = 13$ by $4$ and simplify.

Divide each term in $4 x = 13$ by $4$ .

$\frac{4 x}{4} = \frac{13}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{13}{4}$

Divide $x$ by $1$ .

$x = \frac{13}{4}$

Set $\ln ((2 x - 7) (x - 3))$ equal to $0$ and solve for $x$ .

Set $\ln ((2 x - 7) (x - 3))$ equal to $0$ .

$\ln ((2 x - 7) (x - 3)) = 0$

Solve $\ln ((2 x - 7) (x - 3)) = 0$ for $x$ .

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln ((2 x - 7) (x - 3))} = e^{0}$

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

$e^{0} = (2 x - 7) (x - 3)$

Solve for $x$ .

Rewrite the equation as $(2 x - 7) (x - 3) = e^{0}$ .

$(2 x - 7) (x - 3) = e^{0}$

Simplify $(2 x - 7) (x - 3)$ .

Expand $(2 x - 7) (x - 3)$ using the FOIL Method.

Apply the distributive property.

$2 x (x - 3) - 7 (x - 3) = e^{0}$

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 3 - 7 (x - 3) = e^{0}$

Apply the distributive property.

$2 x \cdot x + 2 x \cdot - 3 - 7 x - 7 \cdot - 3 = e^{0}$

Simplify and combine like terms.

Simplify each term.

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

Move $x$ .

$2 (x \cdot x) + 2 x \cdot - 3 - 7 x - 7 \cdot - 3 = e^{0}$

Multiply $x$ by $x$ .

$2 x^{2} + 2 x \cdot - 3 - 7 x - 7 \cdot - 3 = e^{0}$

Multiply $- 3$ by $2$ .

$2 x^{2} - 6 x - 7 x - 7 \cdot - 3 = e^{0}$

Multiply $- 7$ by $- 3$ .

$2 x^{2} - 6 x - 7 x + 21 = e^{0}$

Subtract $7 x$ from $- 6 x$ .

$2 x^{2} - 13 x + 21 = e^{0}$

Anything raised to $0$ is $1$ .

$2 x^{2} - 13 x + 21 = 1$

Subtract $1$ from both sides of the equation.

$2 x^{2} - 13 x + 21 - 1 = 0$

Subtract $1$ from $21$ .

$2 x^{2} - 13 x + 20 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 20 = 40$ and whose sum is $b = - 13$ .

Factor $- 13$ out of $- 13 x$ .

$2 x^{2} - 13 x + 20 = 0$

Rewrite $- 13$ as $- 5$ plus $- 8$

$2 x^{2} + (- 5 - 8) x + 20 = 0$

Apply the distributive property.

$2 x^{2} - 5 x - 8 x + 20 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} - 5 x) - 8 x + 20 = 0$

Factor out the greatest common factor (GCF) from each group.

$x (2 x - 5) - 4 (2 x - 5) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x - 5$ .

$(2 x - 5) (x - 4) = 0$

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

$2 x - 5 = 0$

$x - 4 = 0$

Set $2 x - 5$ equal to $0$ and solve for $x$ .

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Solve $2 x - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$2 x = 5$

Divide each term in $2 x = 5$ by $2$ and simplify.

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Divide $x$ by $1$ .

$x = \frac{5}{2}$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $(2 x - 5) (x - 4) = 0$ true.

$x = \frac{5}{2}, 4$

The final solution is all the values that make $(4 x - 13) \ln ((2 x - 7) (x - 3)) = 0$ true.

$x = \frac{13}{4}, \frac{5}{2}, 4$

Exclude the solutions that do not make $(4 x - 13) \ln ((2 x - 7) (x - 3)) = 0$ true.

$x = \frac{5}{2}, 4$

Find the Roots (Zeros) f(x)=(4x-13) natural log of (2x-7)(x-3)

Related Questions

Support LCMGCF.com by disabling your adblocker.