Solve for x log of x+1+ log of x-3 = log of 6x^2-6

$\log (x + 1) + \log (x - 3) = \log (6 x^{2} - 6)$

Simplify the left side.

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

$\log ((x + 1) (x - 3)) = \log (6 x^{2} - 6)$

Expand $(x + 1) (x - 3)$ using the FOIL Method.

Apply the distributive property.

$\log (x (x - 3) + 1 (x - 3)) = \log (6 x^{2} - 6)$

Apply the distributive property.

$\log (x \cdot x + x \cdot - 3 + 1 (x - 3)) = \log (6 x^{2} - 6)$

Apply the distributive property.

$\log (x \cdot x + x \cdot - 3 + 1 x + 1 \cdot - 3) = \log (6 x^{2} - 6)$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\log (x^{2} + x \cdot - 3 + 1 x + 1 \cdot - 3) = \log (6 x^{2} - 6)$

Move $- 3$ to the left of $x$ .

$\log (x^{2} - 3 \cdot x + 1 x + 1 \cdot - 3) = \log (6 x^{2} - 6)$

Multiply $x$ by $1$ .

$\log (x^{2} - 3 x + x + 1 \cdot - 3) = \log (6 x^{2} - 6)$

Multiply $- 3$ by $1$ .

$\log (x^{2} - 3 x + x - 3) = \log (6 x^{2} - 6)$

Add $- 3 x$ and $x$ .

$\log (x^{2} - 2 x - 3) = \log (6 x^{2} - 6)$

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

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

Solve for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $6 x^{2}$ from both sides of the equation.

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

Subtract $6 x^{2}$ from $x^{2}$ .

$- 5 x^{2} - 2 x - 3 = - 6$

Add $6$ to both sides of the equation.

$- 5 x^{2} - 2 x - 3 + 6 = 0$

Add $- 3$ and $6$ .

$- 5 x^{2} - 2 x + 3 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 5 x^{2} - 2 x + 3$ .

Factor $- 1$ out of $- 5 x^{2}$ .

$- (5 x^{2}) - 2 x + 3 = 0$

Factor $- 1$ out of $- 2 x$ .

$- (5 x^{2}) - (2 x) + 3 = 0$

Rewrite $3$ as $- 1 (- 3)$ .

$- (5 x^{2}) - (2 x) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (5 x^{2}) - (2 x)$ .

$- (5 x^{2} + 2 x) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (5 x^{2} + 2 x) - 1 (- 3)$ .

$- (5 x^{2} + 2 x - 3) = 0$

Factor.

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 $a \cdot c = 5 \cdot - 3 = - 15$ and whose sum is $b = 2$ .

Factor $2$ out of $2 x$ .

$- (5 x^{2} + 2 (x) - 3) = 0$

Rewrite $2$ as $- 3$ plus $5$

$- (5 x^{2} + (- 3 + 5) x - 3) = 0$

Apply the distributive property.

$- (5 x^{2} - 3 x + 5 x - 3) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((5 x^{2} - 3 x) + 5 x - 3) = 0$

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

$- (x (5 x - 3) + 1 (5 x - 3)) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 x - 3$ .

$- ((5 x - 3) (x + 1)) = 0$

Remove unnecessary parentheses.

$- (5 x - 3) (x + 1) = 0$

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

$5 x - 3 = 0$

$x + 1 = 0$

Set $5 x - 3$ equal to $0$ and solve for $x$ .

Set $5 x - 3$ equal to $0$ .

$5 x - 3 = 0$

Solve $5 x - 3 = 0$ for $x$ .

Add $3$ to both sides of the equation.

$5 x = 3$

Divide each term in $5 x = 3$ by $5$ and simplify.

Divide each term in $5 x = 3$ by $5$ .

$\frac{5 x}{5} = \frac{3}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{3}{5}$

Divide $x$ by $1$ .

$x = \frac{3}{5}$

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 $- (5 x - 3) (x + 1) = 0$ true.

$x = \frac{3}{5}, - 1$

Exclude the solutions that do not make $\log (x + 1) + \log (x - 3) = \log (6 x^{2} - 6)$ true.

$No solution$

Solve for x log of x+1+ log of x-3 = log of 6x^2-6

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