Calculus Examples

$\log_{2} (x + 1) + \log_{2} (x - 3) = 5$

Step 1

Simplify the left side.

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Step 1.1

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

$\log_{2} ((x + 1) (x - 3)) = 5$

Step 1.2

Expand $(x + 1) (x - 3)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\log_{2} (x (x - 3) + 1 (x - 3)) = 5$

Step 1.2.2

Apply the distributive property.

$\log_{2} (x \cdot x + x \cdot - 3 + 1 (x - 3)) = 5$

Step 1.2.3

Apply the distributive property.

$\log_{2} (x \cdot x + x \cdot - 3 + 1 x + 1 \cdot - 3) = 5$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply $x$ by $x$ .

$\log_{2} (x^{2} + x \cdot - 3 + 1 x + 1 \cdot - 3) = 5$

Step 1.3.1.2

Move $- 3$ to the left of $x$ .

$\log_{2} (x^{2} - 3 \cdot x + 1 x + 1 \cdot - 3) = 5$

Step 1.3.1.3

Multiply $x$ by $1$ .

$\log_{2} (x^{2} - 3 x + x + 1 \cdot - 3) = 5$

Step 1.3.1.4

Multiply $- 3$ by $1$ .

$\log_{2} (x^{2} - 3 x + x - 3) = 5$

Step 1.3.2

Add $- 3 x$ and $x$ .

$\log_{2} (x^{2} - 2 x - 3) = 5$

Step 2

Rewrite $\log_{2} (x^{2} - 2 x - 3) = 5$ 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 $b^{y} = x$ .

$2^{5} = x^{2} - 2 x - 3$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $x^{2} - 2 x - 3 = 2^{5}$ .

$x^{2} - 2 x - 3 = 2^{5}$

Step 3.2

Raise $2$ to the power of $5$ .

$x^{2} - 2 x - 3 = 32$

Step 3.3

Subtract $32$ from both sides of the equation.

$x^{2} - 2 x - 3 - 32 = 0$

Step 3.4

Subtract $32$ from $- 3$ .

$x^{2} - 2 x - 35 = 0$

Step 3.5

Factor $x^{2} - 2 x - 35$ using the AC method.

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Step 3.5.1

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 $- 35$ and whose sum is $- 2$ .

$- 7, 5$

Step 3.5.2

Write the factored form using these integers.

$(x - 7) (x + 5) = 0$

Step 3.6

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

$x - 7 = 0$

$x + 5 = 0$

Step 3.7

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 3.7.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 3.7.2

Add $7$ to both sides of the equation.

$x = 7$

Step 3.8

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 3.8.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 3.8.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 3.9

The final solution is all the values that make $(x - 7) (x + 5) = 0$ true.

$x = 7, - 5$

Step 4

Exclude the solutions that do not make $\log_{2} (x + 1) + \log_{2} (x - 3) = 5$ true.

$x = 7$