Solve for x log base 3 of x+ log base 3 of x+1 = log base 3 of 6

$\log_{3} (x) + \log_{3} (x + 1) = \log_{3} (6)$

Simplify the left side.

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

$\log_{3} (x (x + 1)) = \log_{3} (6)$

Apply the distributive property.

$\log_{3} (x \cdot x + x \cdot 1) = \log_{3} (6)$

Simplify the expression.

Multiply $x$ by $x$ .

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

Multiply $x$ by $1$ .

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

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

$x^{2} + x = 6$

Solve for $x$ .

Subtract $6$ from both sides of the equation.

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

Factor $x^{2} + x - 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $1$ .

$- 2, 3$

Write the factored form using these integers.

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

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

$x - 2 = 0$

$x + 3 = 0$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = 2, - 3$

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

$x = 2$

Solve for x log base 3 of x+ log base 3 of x+1 = log base 3 of 6

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