Algebra Examples

$2 \log_{3} (6 x) - \log_{3} (4 x) = 2 \log_{3} (x + 2)$

Step 1

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

$2 \log_{3} (6 x) - \log_{3} (4 x) - 2 \log_{3} (x + 2) = 0$

Step 2

Simplify the left side.

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Step 2.1

Simplify $2 \log_{3} (6 x) - \log_{3} (4 x) - 2 \log_{3} (x + 2)$ .

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Step 2.1.1

Simplify each term.

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Step 2.1.1.1

Simplify $2 \log_{3} (6 x)$ by moving $2$ inside the logarithm.

$\log_{3} ({(6 x)}^{2}) - \log_{3} (4 x) - 2 \log_{3} (x + 2) = 0$

Step 2.1.1.2

Apply the product rule to $6 x$ .

$\log_{3} (6^{2} x^{2}) - \log_{3} (4 x) - 2 \log_{3} (x + 2) = 0$

Step 2.1.1.3

Raise $6$ to the power of $2$ .

$\log_{3} (36 x^{2}) - \log_{3} (4 x) - 2 \log_{3} (x + 2) = 0$

Step 2.1.1.4

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

$\log_{3} (36 x^{2}) - \log_{3} (4 x) - \log_{3} ({(x + 2)}^{2}) = 0$

Step 2.1.2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{3} (\frac{36 x^{2}}{4 x}) - \log_{3} ({(x + 2)}^{2}) = 0$

Step 2.1.3

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{3} (\frac{\frac{36 x^{2}}{4 x}}{{(x + 2)}^{2}}) = 0$

Step 2.1.4

Cancel the common factor of $36$ and $4$ .

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Step 2.1.4.1

Factor $4$ out of $36 x^{2}$ .

$\log_{3} (\frac{\frac{4 (9 x^{2})}{4 x}}{{(x + 2)}^{2}}) = 0$

Step 2.1.4.2

Cancel the common factors.

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Step 2.1.4.2.1

Factor $4$ out of $4 x$ .

$\log_{3} (\frac{\frac{4 (9 x^{2})}{4 (x)}}{{(x + 2)}^{2}}) = 0$

Step 2.1.4.2.2

Cancel the common factor.

$\log_{3} (\frac{\frac{4 (9 x^{2})}{4 x}}{{(x + 2)}^{2}}) = 0$

Step 2.1.4.2.3

Rewrite the expression.

$\log_{3} (\frac{\frac{9 x^{2}}{x}}{{(x + 2)}^{2}}) = 0$

Step 2.1.5

Cancel the common factor of $x^{2}$ and $x$ .

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Step 2.1.5.1

Factor $x$ out of $9 x^{2}$ .

$\log_{3} (\frac{\frac{x (9 x)}{x}}{{(x + 2)}^{2}}) = 0$

Step 2.1.5.2

Cancel the common factors.

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Step 2.1.5.2.1

Raise $x$ to the power of $1$ .

$\log_{3} (\frac{\frac{x (9 x)}{x^{1}}}{{(x + 2)}^{2}}) = 0$

Step 2.1.5.2.2

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

$\log_{3} (\frac{\frac{x (9 x)}{x \cdot 1}}{{(x + 2)}^{2}}) = 0$

Step 2.1.5.2.3

Cancel the common factor.

$\log_{3} (\frac{\frac{x (9 x)}{x \cdot 1}}{{(x + 2)}^{2}}) = 0$

Step 2.1.5.2.4

Rewrite the expression.

$\log_{3} (\frac{\frac{9 x}{1}}{{(x + 2)}^{2}}) = 0$

Step 2.1.5.2.5

Divide $9 x$ by $1$ .

$\log_{3} (\frac{9 x}{{(x + 2)}^{2}}) = 0$

Step 3

Rewrite $\log_{3} (\frac{9 x}{{(x + 2)}^{2}}) = 0$ 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 $b^{y} = x$ .

$3^{0} = \frac{9 x}{{(x + 2)}^{2}}$

Step 4

Cross multiply to remove the fraction.

$9 x = 3^{0} ({(x + 2)}^{2})$

Step 5

Simplify $3^{0} ({(x + 2)}^{2})$ .

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Step 5.1

Simplify the expression.

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Step 5.1.1

Anything raised to $0$ is $1$ .

$9 x = 1 {(x + 2)}^{2}$

Step 5.1.2

Multiply ${(x + 2)}^{2}$ by $1$ .

$9 x = {(x + 2)}^{2}$

Step 5.1.3

Rewrite ${(x + 2)}^{2}$ as $(x + 2) (x + 2)$ .

$9 x = (x + 2) (x + 2)$

Step 5.2

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

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Step 5.2.1

Apply the distributive property.

$9 x = x (x + 2) + 2 (x + 2)$

Step 5.2.2

Apply the distributive property.

$9 x = x \cdot x + x \cdot 2 + 2 (x + 2)$

Step 5.2.3

Apply the distributive property.

$9 x = x \cdot x + x \cdot 2 + 2 x + 2 \cdot 2$

Step 5.3

Simplify and combine like terms.

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Step 5.3.1

Simplify each term.

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Step 5.3.1.1

Multiply $x$ by $x$ .

$9 x = x^{2} + x \cdot 2 + 2 x + 2 \cdot 2$

Step 5.3.1.2

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

$9 x = x^{2} + 2 \cdot x + 2 x + 2 \cdot 2$

Step 5.3.1.3

Multiply $2$ by $2$ .

$9 x = x^{2} + 2 x + 2 x + 4$

Step 5.3.2

Add $2 x$ and $2 x$ .

$9 x = x^{2} + 4 x + 4$

Step 6

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

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Step 6.1

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

$9 x - x^{2} = 4 x + 4$

Step 6.2

Subtract $4 x$ from both sides of the equation.

$9 x - x^{2} - 4 x = 4$

Step 6.3

Subtract $4 x$ from $9 x$ .

$- x^{2} + 5 x = 4$

Step 7

Factor $x$ out of $- x^{2} + 5 x$ .

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Step 7.1

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

$x (- x) + 5 x = 4$

Step 7.2

Factor $x$ out of $5 x$ .

$x (- x) + x \cdot 5 = 4$

Step 7.3

Factor $x$ out of $x (- x) + x \cdot 5$ .

$x (- x + 5) = 4$

Step 8

Simplify $x (- x + 5)$ .

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Step 8.1

Simplify by multiplying through.

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Step 8.1.1

Apply the distributive property.

$x (- x) + x \cdot 5 = 4$

Step 8.1.2

Reorder.

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Step 8.1.2.1

Rewrite using the commutative property of multiplication.

$- x \cdot x + x \cdot 5 = 4$

Step 8.1.2.2

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

$- x \cdot x + 5 \cdot x = 4$

Step 8.2

Multiply $x$ by $x$ by adding the exponents.

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Step 8.2.1

Move $x$ .

$- (x \cdot x) + 5 \cdot x = 4$

Step 8.2.2

Multiply $x$ by $x$ .

$- x^{2} + 5 \cdot x = 4$

$- x^{2} + 5 x = 4$

Step 9

Subtract $4$ from both sides of the equation.

$- x^{2} + 5 x - 4 = 0$

Step 10

Factor the left side of the equation.

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Step 10.1

Factor $- 1$ out of $- x^{2} + 5 x - 4$ .

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Step 10.1.1

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

$- (x^{2}) + 5 x - 4 = 0$

Step 10.1.2

Factor $- 1$ out of $5 x$ .

$- (x^{2}) - (- 5 x) - 4 = 0$

Step 10.1.3

Rewrite $- 4$ as $- 1 (4)$ .

$- (x^{2}) - (- 5 x) - 1 \cdot 4 = 0$

Step 10.1.4

Factor $- 1$ out of $- (x^{2}) - (- 5 x)$ .

$- (x^{2} - 5 x) - 1 \cdot 4 = 0$

Step 10.1.5

Factor $- 1$ out of $- (x^{2} - 5 x) - 1 (4)$ .

$- (x^{2} - 5 x + 4) = 0$

Step 10.2

Factor.

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Step 10.2.1

Factor $x^{2} - 5 x + 4$ using the AC method.

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Step 10.2.1.1

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 $4$ and whose sum is $- 5$ .

$- 4, - 1$

Step 10.2.1.2

Write the factored form using these integers.

$- ((x - 4) (x - 1)) = 0$

Step 10.2.2

Remove unnecessary parentheses.

$- (x - 4) (x - 1) = 0$

Step 11

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

$x - 4 = 0$

$x - 1 = 0$

Step 12

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 12.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 12.2

Add $4$ to both sides of the equation.

$x = 4$

Step 13

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

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Step 13.1

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

$x - 1 = 0$

Step 13.2

Add $1$ to both sides of the equation.

$x = 1$

Step 14

The final solution is all the values that make $- (x - 4) (x - 1) = 0$ true.

$x = 4, 1$