Solve for x log base 3 of x^2+2x = log base 3 of x+2

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

Step 1

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

$x^{2} + 2 x = x + 2$

Step 2

Solve for $x$ .

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

Subtract $x$ from both sides of the equation.

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

Subtract $x$ from $2 x$ .

$x^{2} + x = 2$

Subtract $2$ from both sides of the equation.

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

Factor $x^{2} + x - 2$ 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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

$(x - 1) (x + 2) = 0$

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

$x - 1 = 0$

$x + 2 = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 1) (x + 2) = 0$ true.

$x = 1, - 2$

Step 3

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

$x = 1$

Solve for x log base 3 of x^2+2x = log base 3 of x+2

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