Precalculus Examples

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

Step 1

Simplify the left side.

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Step 1.1

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

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

Step 1.2

Apply the distributive property.

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

Step 1.3

Simplify the expression.

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Step 1.3.1

Multiply $x$ by $x$ .

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

Step 1.3.2

Move $2$ to the left of $x$ .

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

Step 2

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

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

Step 3

Solve for $x$ .

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Step 3.1

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

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Step 3.1.1

Subtract $x$ from both sides of the equation.

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

Step 3.1.2

Subtract $x$ from $2 x$ .

$x^{2} + x = 6$

Step 3.2

Subtract $6$ from both sides of the equation.

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

Step 3.3

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

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Step 3.3.1

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 $1$ .

$- 2, 3$

Step 3.3.2

Write the factored form using these integers.

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

Step 3.4

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

$x - 2 = 0$

$x + 3 = 0$

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

Add $2$ to both sides of the equation.

$x = 2$

Step 3.6

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

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Step 3.6.1

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

$x + 3 = 0$

Step 3.6.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 3.7

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

$x = 2, - 3$

Step 4

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

$x = 2$