Algebra Examples

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

Step 1

Simplify $2 \log_{3} (x)$ by moving $2$ inside the logarithm.

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

Step 2

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

$x + 6 = x^{2}$

Step 3

Solve for $x$ .

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Step 3.1

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

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

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

Factor $- 1$ out of $x + 6 - x^{2}$ .

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Step 3.2.1.1

Reorder the expression.

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Step 3.2.1.1.1

Move $6$ .

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

Step 3.2.1.1.2

Reorder $x$ and $- x^{2}$ .

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

Step 3.2.1.2

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

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

Step 3.2.1.3

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 6 = 0$

Step 3.2.1.4

Rewrite $6$ as $- 1 (- 6)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 6 = 0$

Step 3.2.1.5

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

$- (x^{2} - x) - 1 \cdot - 6 = 0$

Step 3.2.1.6

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 6)$ .

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

Step 3.2.2

Factor.

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Step 3.2.2.1

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

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Step 3.2.2.1.1

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 $- 1$ .

$- 3, 2$

Step 3.2.2.1.2

Write the factored form using these integers.

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

Step 3.2.2.2

Remove unnecessary parentheses.

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

Step 3.3

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

$x - 3 = 0$

$x + 2 = 0$

Step 3.4

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

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Step 3.4.1

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

$x - 3 = 0$

Step 3.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 3.5

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

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Step 3.5.1

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

$x + 2 = 0$

Step 3.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 3.6

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

$x = 3, - 2$

Step 4

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

$x = 3$