Precalculus Examples

{log}_{2} (x) + {log}_{2} (x + 2) = {log}_{2} (x + 6)

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

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)

$\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

{log}_{2} (x (x + 2)) = {log}_{2} (x + 6)

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

Step 1.2

Apply the distributive property.

{log}_{2} (x \cdot x + x \cdot 2) = {log}_{2} (x + 6)

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

Step 1.3

Simplify the expression.

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Step 1.3.1

Multiply

x

$x$ by

x

$x$ .

{log}_{2} (x^{2} + x \cdot 2) = {log}_{2} (x + 6)

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

Step 1.3.2

Move

2

$2$ to the left of

x

$x$ .

{log}_{2} (x^{2} + 2 x) = {log}_{2} (x + 6)

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

{log}_{2} (x^{2} + 2 x) = {log}_{2} (x + 6)

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

{log}_{2} (x^{2} + 2 x) = {log}_{2} (x + 6)

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

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

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

Step 3

Solve for

x

$x$ .

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Step 3.1

Move all terms containing

x

$x$ to the left side of the equation.

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Step 3.1.1

Subtract

x

$x$ from both sides of the equation.

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

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

Step 3.1.2

Subtract

x

$x$ from

2 x

$2 x$ .

x^{2} + x = 6

$x^{2} + x = 6$

x^{2} + x = 6

$x^{2} + x = 6$

Step 3.2

Subtract

6

$6$ from both sides of the equation.

x^{2} + x - 6 = 0

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

Step 3.3

Factor

x^{2} + x - 6

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

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Step 3.3.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

- 6

$- 6$ and whose sum is

1

$1$ .

- 2, 3

$- 2, 3$

Step 3.3.2

Write the factored form using these integers.

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

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

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

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

Step 3.4

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

x - 2 = 0

$x - 2 = 0$

x + 3 = 0

$x + 3 = 0$

Step 3.5

Set

x - 2

$x - 2$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.5.1

Set

x - 2

$x - 2$ equal to

0

$0$ .

x - 2 = 0

$x - 2 = 0$

Step 3.5.2

Add

2

$2$ to both sides of the equation.

x = 2

$x = 2$

x = 2

$x = 2$

Step 3.6

Set

x + 3

$x + 3$ equal to

0

$0$ and solve for

x

$x$ .

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Step 3.6.1

Set

x + 3

$x + 3$ equal to

0

$0$ .

x + 3 = 0

$x + 3 = 0$

Step 3.6.2

Subtract

3

$3$ from both sides of the equation.

x = - 3

$x = - 3$

x = - 3

$x = - 3$

Step 3.7

The final solution is all the values that make

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

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

x = 2, - 3

$x = 2, - 3$

x = 2, - 3

$x = 2, - 3$

Step 4

Exclude the solutions that do not make

{log}_{2} (x) + {log}_{2} (x + 2) = {log}_{2} (x + 6)

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

x = 2

$x = 2$