Solve for x log base 7 of 2x^2-7- log base 7 of 3x+2=0

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

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{7} (\frac{2 x^{2} - 7}{3 x + 2}) = 0$

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

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

Cross multiply to remove the fraction.

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

Simplify $7^{0} (3 x + 2)$ .

Anything raised to $0$ is $1$ .

$2 x^{2} - 7 = 1 (3 x + 2)$

Multiply $3 x + 2$ by $1$ .

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

Subtract $3 x$ from both sides of the equation.

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

Move all terms not containing $x$ to the right side of the equation.

Add $7$ to both sides of the equation.

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

Add $2$ and $7$ .

$2 x^{2} - 3 x = 9$

Subtract $9$ from both sides of the equation.

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

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 $a \cdot c = 2 \cdot - 9 = - 18$ and whose sum is $b = - 3$ .

Factor $- 3$ out of $- 3 x$ .

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

Rewrite $- 3$ as $3$ plus $- 6$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$x (2 x + 3) - 3 (2 x + 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x + 3$ .

$(2 x + 3) (x - 3) = 0$

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

$2 x + 3 = 0$

$x - 3 = 0$

Set $2 x + 3$ equal to $0$ and solve for $x$ .

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Solve $2 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Divide each term in $2 x = - 3$ by $2$ and simplify.

Divide each term in $2 x = - 3$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2}$

Divide $x$ by $1$ .

$x = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

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.

$x = 3$

The final solution is all the values that make $(2 x + 3) (x - 3) = 0$ true.

$x = - \frac{3}{2}, 3$

Exclude the solutions that do not make $\log_{7} (2 x^{2} - 7) - \log_{7} (3 x + 2) = 0$ true.

$x = 3$

Solve for x log base 7 of 2x^2-7- log base 7 of 3x+2=0

Related Questions

Support LCMGCF.com by disabling your adblocker.