Algebra Examples

$\log_{2} (4 x^{2} + 7) - \log_{2} (2 x + 5) = 3$

Step 1

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

$\log_{2} (\frac{4 x^{2} + 7}{2 x + 5}) = 3$

Step 2

Rewrite $\log_{2} (\frac{4 x^{2} + 7}{2 x + 5}) = 3$ 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 $b^{y} = x$ .

$2^{3} = \frac{4 x^{2} + 7}{2 x + 5}$

Step 3

Cross multiply to remove the fraction.

$4 x^{2} + 7 = 2^{3} (2 x + 5)$

Step 4

Simplify $2^{3} (2 x + 5)$ .

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Step 4.1

Raise $2$ to the power of $3$ .

$4 x^{2} + 7 = 8 (2 x + 5)$

Step 4.2

Apply the distributive property.

$4 x^{2} + 7 = 8 (2 x) + 8 \cdot 5$

Step 4.3

Multiply.

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Step 4.3.1

Multiply $2$ by $8$ .

$4 x^{2} + 7 = 16 x + 8 \cdot 5$

Step 4.3.2

Multiply $8$ by $5$ .

$4 x^{2} + 7 = 16 x + 40$

Step 5

Subtract $16 x$ from both sides of the equation.

$4 x^{2} + 7 - 16 x = 40$

Step 6

Factor by grouping.

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Step 6.1

Reorder terms.

$4 x^{2} - 16 x + 7 = 40$

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 $a \cdot c = 4 \cdot 7 = 28$ and whose sum is $b = - 16$ .

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Step 6.2.1

Factor $- 16$ out of $- 16 x$ .

$4 x^{2} - 16 x + 7 = 40$

Step 6.2.2

Rewrite $- 16$ as $- 2$ plus $- 14$

$4 x^{2} + (- 2 - 14) x + 7 = 40$

Step 6.2.3

Apply the distributive property.

$4 x^{2} - 2 x - 14 x + 7 = 40$

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.

$(4 x^{2} - 2 x) - 14 x + 7 = 40$

Step 6.3.2

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

$2 x (2 x - 1) - 7 (2 x - 1) = 40$

Step 6.4

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (2 x - 7) = 40$

Step 7

Simplify $(2 x - 1) (2 x - 7)$ .

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Step 7.1

Expand $(2 x - 1) (2 x - 7)$ using the FOIL Method.

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Step 7.1.1

Apply the distributive property.

$2 x (2 x - 7) - 1 (2 x - 7) = 40$

Step 7.1.2

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 7 - 1 (2 x - 7) = 40$

Step 7.1.3

Apply the distributive property.

$2 x (2 x) + 2 x \cdot - 7 - 1 (2 x) - 1 \cdot - 7 = 40$

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

Rewrite using the commutative property of multiplication.

$2 \cdot 2 x \cdot x + 2 x \cdot - 7 - 1 (2 x) - 1 \cdot - 7 = 40$

Step 7.2.1.2

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

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Step 7.2.1.2.1

Move $x$ .

$2 \cdot 2 (x \cdot x) + 2 x \cdot - 7 - 1 (2 x) - 1 \cdot - 7 = 40$

Step 7.2.1.2.2

Multiply $x$ by $x$ .

$2 \cdot 2 x^{2} + 2 x \cdot - 7 - 1 (2 x) - 1 \cdot - 7 = 40$

Step 7.2.1.3

Multiply $2$ by $2$ .

$4 x^{2} + 2 x \cdot - 7 - 1 (2 x) - 1 \cdot - 7 = 40$

Step 7.2.1.4

Multiply $- 7$ by $2$ .

$4 x^{2} - 14 x - 1 (2 x) - 1 \cdot - 7 = 40$

Step 7.2.1.5

Multiply $2$ by $- 1$ .

$4 x^{2} - 14 x - 2 x - 1 \cdot - 7 = 40$

Step 7.2.1.6

Multiply $- 1$ by $- 7$ .

$4 x^{2} - 14 x - 2 x + 7 = 40$

Step 7.2.2

Subtract $2 x$ from $- 14 x$ .

$4 x^{2} - 16 x + 7 = 40$

Step 8

Subtract $40$ from both sides of the equation.

$4 x^{2} - 16 x + 7 - 40 = 0$

Step 9

Subtract $40$ from $7$ .

$4 x^{2} - 16 x - 33 = 0$

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 $a \cdot c = 4 \cdot - 33 = - 132$ and whose sum is $b = - 16$ .

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Step 10.1.1

Factor $- 16$ out of $- 16 x$ .

$4 x^{2} - 16 x - 33 = 0$

Step 10.1.2

Rewrite $- 16$ as $6$ plus $- 22$

$4 x^{2} + (6 - 22) x - 33 = 0$

Step 10.1.3

Apply the distributive property.

$4 x^{2} + 6 x - 22 x - 33 = 0$

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.

$(4 x^{2} + 6 x) - 22 x - 33 = 0$

Step 10.2.2

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

$2 x (2 x + 3) - 11 (2 x + 3) = 0$

Step 10.3

Factor the polynomial by factoring out the greatest common factor, $2 x + 3$ .

$(2 x + 3) (2 x - 11) = 0$

Step 11

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

$2 x + 3 = 0$

$2 x - 11 = 0$

Step 12

Set $2 x + 3$ equal to $0$ and solve for $x$ .

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Step 12.1

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Step 12.2

Solve $2 x + 3 = 0$ for $x$ .

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Step 12.2.1

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 12.2.2

Divide each term in $2 x = - 3$ by $2$ and simplify.

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Step 12.2.2.1

Divide each term in $2 x = - 3$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2}$

Step 12.2.2.2

Simplify the left side.

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Step 12.2.2.2.1

Cancel the common factor of $2$ .

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Step 12.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2}$

Step 12.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 3}{2}$

Step 12.2.2.3

Simplify the right side.

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Step 12.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

Step 13

Set $2 x - 11$ equal to $0$ and solve for $x$ .

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Step 13.1

Set $2 x - 11$ equal to $0$ .

$2 x - 11 = 0$

Step 13.2

Solve $2 x - 11 = 0$ for $x$ .

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Step 13.2.1

Add $11$ to both sides of the equation.

$2 x = 11$

Step 13.2.2

Divide each term in $2 x = 11$ by $2$ and simplify.

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Step 13.2.2.1

Divide each term in $2 x = 11$ by $2$ .

$\frac{2 x}{2} = \frac{11}{2}$

Step 13.2.2.2

Simplify the left side.

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Step 13.2.2.2.1

Cancel the common factor of $2$ .

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Step 13.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{11}{2}$

Step 13.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{11}{2}$

Step 14

The final solution is all the values that make $(2 x + 3) (2 x - 11) = 0$ true.

$x = - \frac{3}{2}, \frac{11}{2}$

Step 15

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{3}{2}, \frac{11}{2}$

Decimal Form:

$x = - 1.5, 5.5$

Mixed Number Form:

$x = - 1 \frac{1}{2}, 5 \frac{1}{2}$