Precalculus Examples

$\log_{2} (x + 2) = 2 - \log_{2} (x + 5)$

Step 1

Move all the terms containing a logarithm to the left side of the equation.

$\log_{2} (x + 2) + \log_{2} (x + 5) = 2$

Step 2

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

$\log_{2} ((x + 2) (x + 5)) = 2$

Step 3

Expand $(x + 2) (x + 5)$ using the FOIL Method.

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Step 3.1

Apply the distributive property.

$\log_{2} (x (x + 5) + 2 (x + 5)) = 2$

Step 3.2

Apply the distributive property.

$\log_{2} (x \cdot x + x \cdot 5 + 2 (x + 5)) = 2$

Step 3.3

Apply the distributive property.

$\log_{2} (x \cdot x + x \cdot 5 + 2 x + 2 \cdot 5) = 2$

Step 4

Simplify and combine like terms.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply $x$ by $x$ .

$\log_{2} (x^{2} + x \cdot 5 + 2 x + 2 \cdot 5) = 2$

Step 4.1.2

Move $5$ to the left of $x$ .

$\log_{2} (x^{2} + 5 \cdot x + 2 x + 2 \cdot 5) = 2$

Step 4.1.3

Multiply $2$ by $5$ .

$\log_{2} (x^{2} + 5 x + 2 x + 10) = 2$

Step 4.2

Add $5 x$ and $2 x$ .

$\log_{2} (x^{2} + 7 x + 10) = 2$

Step 5

Rewrite $\log_{2} (x^{2} + 7 x + 10) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$2^{2} = x^{2} + 7 x + 10$

Step 6

Solve for $x$ .

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Step 6.1

Rewrite the equation as $x^{2} + 7 x + 10 = 2^{2}$ .

$x^{2} + 7 x + 10 = 2^{2}$

Step 6.2

Raise $2$ to the power of $2$ .

$x^{2} + 7 x + 10 = 4$

Step 6.3

Subtract $4$ from both sides of the equation.

$x^{2} + 7 x + 10 - 4 = 0$

Step 6.4

Subtract $4$ from $10$ .

$x^{2} + 7 x + 6 = 0$

Step 6.5

Factor $x^{2} + 7 x + 6$ using the AC method.

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Step 6.5.1

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 $7$ .

$1, 6$

Step 6.5.2

Write the factored form using these integers.

$(x + 1) (x + 6) = 0$

Step 6.6

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

$x + 1 = 0$

$x + 6 = 0$

Step 6.7

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

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Step 6.7.1

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

$x + 1 = 0$

Step 6.7.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6.8

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 6.8.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 6.8.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 6.9

The final solution is all the values that make $(x + 1) (x + 6) = 0$ true.

$x = - 1, - 6$

Step 7

Exclude the solutions that do not make $\log_{2} (x + 2) = 2 - \log_{2} (x + 5)$ true.

$x = - 1$