Algebra Examples

$\log_{7} (4 x^{2} - 1) - \log_{7} (3 x + 6) = 0$

Step 1

Simplify the left side.

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Step 1.1

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

$\log_{7} (\frac{4 x^{2} - 1}{3 x + 6}) = 0$

Step 1.2

Simplify the numerator.

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Step 1.2.1

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$\log_{7} (\frac{{(2 x)}^{2} - 1}{3 x + 6}) = 0$

Step 1.2.2

Rewrite $1$ as $1^{2}$ .

$\log_{7} (\frac{{(2 x)}^{2} - 1^{2}}{3 x + 6}) = 0$

Step 1.2.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 1$ .

$\log_{7} (\frac{(2 x + 1) (2 x - 1)}{3 x + 6}) = 0$

Step 1.3

Factor $3$ out of $3 x + 6$ .

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Step 1.3.1

Factor $3$ out of $3 x$ .

$\log_{7} (\frac{(2 x + 1) (2 x - 1)}{3 (x) + 6}) = 0$

Step 1.3.2

Factor $3$ out of $6$ .

$\log_{7} (\frac{(2 x + 1) (2 x - 1)}{3 x + 3 \cdot 2}) = 0$

Step 1.3.3

Factor $3$ out of $3 x + 3 \cdot 2$ .

$\log_{7} (\frac{(2 x + 1) (2 x - 1)}{3 (x + 2)}) = 0$

Step 2

Rewrite $\log_{7} (\frac{(2 x + 1) (2 x - 1)}{3 (x + 2)}) = 0$ 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 $b^{y} = x$ .

$7^{0} = \frac{(2 x + 1) (2 x - 1)}{3 (x + 2)}$

Step 3

Cross multiply to remove the fraction.

$(2 x + 1) (2 x - 1) = 7^{0} (3 (x + 2))$

Step 4

Simplify $7^{0} (3 (x + 2))$ .

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Step 4.1

Simplify the expression.

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Step 4.1.1

Anything raised to $0$ is $1$ .

$(2 x + 1) (2 x - 1) = 1 (3 (x + 2))$

Step 4.1.2

Multiply $3 (x + 2)$ by $1$ .

$(2 x + 1) (2 x - 1) = 3 (x + 2)$

Step 4.2

Apply the distributive property.

$(2 x + 1) (2 x - 1) = 3 x + 3 \cdot 2$

Step 4.3

Multiply $3$ by $2$ .

$(2 x + 1) (2 x - 1) = 3 x + 6$

Step 5

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

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Step 5.1

Subtract $3 x$ from both sides of the equation.

$(2 x + 1) (2 x - 1) - 3 x = 6$

Step 5.2

Simplify each term.

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Step 5.2.1

Expand $(2 x + 1) (2 x - 1)$ using the FOIL Method.

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Step 5.2.1.1

Apply the distributive property.

$2 x (2 x - 1) + 1 (2 x - 1) - 3 x = 6$

Step 5.2.1.2

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 1 + 1 (2 x - 1) - 3 x = 6$

Step 5.2.1.3

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 1 + 1 (2 x) + 1 \cdot - 1 - 3 x = 6$

Step 5.2.2

Combine the opposite terms in $2 x (2 x) + 2 x \cdot - 1 + 1 (2 x) + 1 \cdot - 1$ .

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Step 5.2.2.1

Reorder the factors in the terms $2 x \cdot - 1$ and $1 (2 x)$ .

$2 x (2 x) - 1 \cdot 2 x + 1 \cdot 2 x + 1 \cdot - 1 - 3 x = 6$

Step 5.2.2.2

Add $- 1 \cdot 2 x$ and $1 \cdot 2 x$ .

$2 x (2 x) + 0 + 1 \cdot - 1 - 3 x = 6$

Step 5.2.2.3

Add $2 x (2 x)$ and $0$ .

$2 x (2 x) + 1 \cdot - 1 - 3 x = 6$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Rewrite using the commutative property of multiplication.

$2 \cdot 2 x \cdot x + 1 \cdot - 1 - 3 x = 6$

Step 5.2.3.2

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

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Step 5.2.3.2.1

Move $x$ .

$2 \cdot 2 (x \cdot x) + 1 \cdot - 1 - 3 x = 6$

Step 5.2.3.2.2

Multiply $x$ by $x$ .

$2 \cdot 2 x^{2} + 1 \cdot - 1 - 3 x = 6$

Step 5.2.3.3

Multiply $2$ by $2$ .

$4 x^{2} + 1 \cdot - 1 - 3 x = 6$

Step 5.2.3.4

Multiply $- 1$ by $1$ .

$4 x^{2} - 1 - 3 x = 6$

Step 6

Factor by grouping.

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Step 6.1

Reorder terms.

$4 x^{2} - 3 x - 1 = 6$

Step 6.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot - 1 = - 4$ and whose sum is $b = - 3$ .

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Step 6.2.1

Factor $- 3$ out of $- 3 x$ .

$4 x^{2} - 3 x - 1 = 6$

Step 6.2.2

Rewrite $- 3$ as $1$ plus $- 4$

$4 x^{2} + (1 - 4) x - 1 = 6$

Step 6.2.3

Apply the distributive property.

$4 x^{2} + 1 x - 4 x - 1 = 6$

Step 6.2.4

Multiply $x$ by $1$ .

$4 x^{2} + x - 4 x - 1 = 6$

Step 6.3

Factor out the greatest common factor from each group.

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Step 6.3.1

Group the first two terms and the last two terms.

$(4 x^{2} + x) - 4 x - 1 = 6$

Step 6.3.2

Factor out the greatest common factor (GCF) from each group.

$x (4 x + 1) - (4 x + 1) = 6$

Step 6.4

Factor the polynomial by factoring out the greatest common factor, $4 x + 1$ .

$(4 x + 1) (x - 1) = 6$

Step 7

Simplify $(4 x + 1) (x - 1)$ .

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Step 7.1

Expand $(4 x + 1) (x - 1)$ using the FOIL Method.

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Step 7.1.1

Apply the distributive property.

$4 x (x - 1) + 1 (x - 1) = 6$

Step 7.1.2

Apply the distributive property.

$4 x \cdot x + 4 x \cdot - 1 + 1 (x - 1) = 6$

Step 7.1.3

Apply the distributive property.

$4 x \cdot x + 4 x \cdot - 1 + 1 x + 1 \cdot - 1 = 6$

Step 7.2

Simplify and combine like terms.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

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

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Step 7.2.1.1.1

Move $x$ .

$4 (x \cdot x) + 4 x \cdot - 1 + 1 x + 1 \cdot - 1 = 6$

Step 7.2.1.1.2

Multiply $x$ by $x$ .

$4 x^{2} + 4 x \cdot - 1 + 1 x + 1 \cdot - 1 = 6$

Step 7.2.1.2

Multiply $- 1$ by $4$ .

$4 x^{2} - 4 x + 1 x + 1 \cdot - 1 = 6$

Step 7.2.1.3

Multiply $x$ by $1$ .

$4 x^{2} - 4 x + x + 1 \cdot - 1 = 6$

Step 7.2.1.4

Multiply $- 1$ by $1$ .

$4 x^{2} - 4 x + x - 1 = 6$

Step 7.2.2

Add $- 4 x$ and $x$ .

$4 x^{2} - 3 x - 1 = 6$

Step 8

Subtract $6$ from both sides of the equation.

$4 x^{2} - 3 x - 1 - 6 = 0$

Step 9

Subtract $6$ from $- 1$ .

$4 x^{2} - 3 x - 7 = 0$

Step 10

Factor by grouping.

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Step 10.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot - 7 = - 28$ and whose sum is $b = - 3$ .

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Step 10.1.1

Factor $- 3$ out of $- 3 x$ .

$4 x^{2} - 3 x - 7 = 0$

Step 10.1.2

Rewrite $- 3$ as $4$ plus $- 7$

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

Step 10.1.3

Apply the distributive property.

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

Step 10.2

Factor out the greatest common factor from each group.

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Step 10.2.1

Group the first two terms and the last two terms.

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

Step 10.2.2

Factor out the greatest common factor (GCF) from each group.

$4 x (x + 1) - 7 (x + 1) = 0$

Step 10.3

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

$(x + 1) (4 x - 7) = 0$

Step 11

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

$x + 1 = 0$

$4 x - 7 = 0$

Step 12

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

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Step 12.1

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

$x + 1 = 0$

Step 12.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 13

Set $4 x - 7$ equal to $0$ and solve for $x$ .

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Step 13.1

Set $4 x - 7$ equal to $0$ .

$4 x - 7 = 0$

Step 13.2

Solve $4 x - 7 = 0$ for $x$ .

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Step 13.2.1

Add $7$ to both sides of the equation.

$4 x = 7$

Step 13.2.2

Divide each term in $4 x = 7$ by $4$ and simplify.

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Step 13.2.2.1

Divide each term in $4 x = 7$ by $4$ .

$\frac{4 x}{4} = \frac{7}{4}$

Step 13.2.2.2

Simplify the left side.

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Step 13.2.2.2.1

Cancel the common factor of $4$ .

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Step 13.2.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{7}{4}$

Step 13.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{7}{4}$

Step 14

The final solution is all the values that make $(x + 1) (4 x - 7) = 0$ true.

$x = - 1, \frac{7}{4}$

Step 15

The result can be shown in multiple forms.

Exact Form:

$x = - 1, \frac{7}{4}$

Decimal Form:

$x = - 1, 1.75$

Mixed Number Form:

$x = - 1, 1 \frac{3}{4}$