Finite Math Examples

$3 = \log_{b} (\frac{27}{64})$

Step 1

Rewrite the equation as $\log_{b} (\frac{27}{64}) = 3$ .

$\log_{b} (\frac{27}{64}) = 3$

Step 2

Rewrite $\log_{b} (\frac{27}{64}) = 3$ 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 $b^{y} = x$ .

$b^{3} = \frac{27}{64}$

Step 3

Solve for $b$ .

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Step 3.1

Subtract $\frac{27}{64}$ from both sides of the equation.

$b^{3} - \frac{27}{64} = 0$

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

Rewrite $\frac{27}{64}$ as ${(\frac{3}{4})}^{3}$ .

$b^{3} - {(\frac{3}{4})}^{3} = 0$

Step 3.2.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = b$ and $b = \frac{3}{4}$ .

$(b - \frac{3}{4}) (b^{2} + b (\frac{3}{4}) + {(\frac{3}{4})}^{2}) = 0$

Step 3.2.3

Simplify.

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Step 3.2.3.1

Combine $b$ and $\frac{3}{4}$ .

$(b - \frac{3}{4}) (b^{2} + \frac{b \cdot 3}{4} + {(\frac{3}{4})}^{2}) = 0$

Step 3.2.3.2

Move $3$ to the left of $b$ .

$(b - \frac{3}{4}) (b^{2} + \frac{3 \cdot b}{4} + {(\frac{3}{4})}^{2}) = 0$

Step 3.2.3.3

Multiply $3$ by $b$ .

$(b - \frac{3}{4}) (b^{2} + \frac{3 b}{4} + {(\frac{3}{4})}^{2}) = 0$

Step 3.2.3.4

Apply the product rule to $\frac{3}{4}$ .

$(b - \frac{3}{4}) (b^{2} + \frac{3 b}{4} + \frac{3^{2}}{4^{2}}) = 0$

Step 3.2.3.5

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

$(b - \frac{3}{4}) (b^{2} + \frac{3 b}{4} + \frac{9}{4^{2}}) = 0$

Step 3.2.3.6

Raise $4$ to the power of $2$ .

$(b - \frac{3}{4}) (b^{2} + \frac{3 b}{4} + \frac{9}{16}) = 0$

Step 3.3

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

$b - \frac{3}{4} = 0$

$b^{2} + \frac{3 b}{4} + \frac{9}{16} = 0$

Step 3.4

Set $b - \frac{3}{4}$ equal to $0$ and solve for $b$ .

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Step 3.4.1

Set $b - \frac{3}{4}$ equal to $0$ .

$b - \frac{3}{4} = 0$

Step 3.4.2

Add $\frac{3}{4}$ to both sides of the equation.

$b = \frac{3}{4}$

Step 3.5

Set $b^{2} + \frac{3 b}{4} + \frac{9}{16}$ equal to $0$ and solve for $b$ .

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Step 3.5.1

Set $b^{2} + \frac{3 b}{4} + \frac{9}{16}$ equal to $0$ .

$b^{2} + \frac{3 b}{4} + \frac{9}{16} = 0$

Step 3.5.2

Solve $b^{2} + \frac{3 b}{4} + \frac{9}{16} = 0$ for $b$ .

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Step 3.5.2.1

Multiply through by the least common denominator $16$ , then simplify.

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Step 3.5.2.1.1

Apply the distributive property.

$16 b^{2} + 16 (\frac{3 b}{4}) + 16 (\frac{9}{16}) = 0$

Step 3.5.2.1.2

Simplify.

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Step 3.5.2.1.2.1

Cancel the common factor of $4$ .

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Step 3.5.2.1.2.1.1

Factor $4$ out of $16$ .

$16 b^{2} + 4 (4) (\frac{3 b}{4}) + 16 (\frac{9}{16}) = 0$

Step 3.5.2.1.2.1.2

Cancel the common factor.

$16 b^{2} + 4 \cdot (4 (\frac{3 b}{4})) + 16 (\frac{9}{16}) = 0$

Step 3.5.2.1.2.1.3

Rewrite the expression.

$16 b^{2} + 4 (3 b) + 16 (\frac{9}{16}) = 0$

Step 3.5.2.1.2.2

Multiply $3$ by $4$ .

$16 b^{2} + 12 b + 16 (\frac{9}{16}) = 0$

Step 3.5.2.1.2.3

Cancel the common factor of $16$ .

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Step 3.5.2.1.2.3.1

Cancel the common factor.

$16 b^{2} + 12 b + 16 (\frac{9}{16}) = 0$

Step 3.5.2.1.2.3.2

Rewrite the expression.

$16 b^{2} + 12 b + 9 = 0$

Step 3.5.2.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.5.2.3

Substitute the values $a = 16$ , $b = 12$ , and $c = 9$ into the quadratic formula and solve for $b$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (16 \cdot 9)}}{2 \cdot 16}$

Step 3.5.2.4

Simplify.

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Step 3.5.2.4.1

Simplify the numerator.

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Step 3.5.2.4.1.1

Raise $12$ to the power of $2$ .

$b = \frac{- 12 \pm \sqrt{144 - 4 \cdot 16 \cdot 9}}{2 \cdot 16}$

Step 3.5.2.4.1.2

Multiply $- 4 \cdot 16 \cdot 9$ .

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Step 3.5.2.4.1.2.1

Multiply $- 4$ by $16$ .

$b = \frac{- 12 \pm \sqrt{144 - 64 \cdot 9}}{2 \cdot 16}$

Step 3.5.2.4.1.2.2

Multiply $- 64$ by $9$ .

$b = \frac{- 12 \pm \sqrt{144 - 576}}{2 \cdot 16}$

Step 3.5.2.4.1.3

Subtract $576$ from $144$ .

$b = \frac{- 12 \pm \sqrt{- 432}}{2 \cdot 16}$

Step 3.5.2.4.1.4

Rewrite $- 432$ as $- 1 (432)$ .

$b = \frac{- 12 \pm \sqrt{- 1 \cdot 432}}{2 \cdot 16}$

Step 3.5.2.4.1.5

Rewrite $\sqrt{- 1 (432)}$ as $\sqrt{- 1} \cdot \sqrt{432}$ .

$b = \frac{- 12 \pm \sqrt{- 1} \cdot \sqrt{432}}{2 \cdot 16}$

Step 3.5.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$b = \frac{- 12 \pm i \cdot \sqrt{432}}{2 \cdot 16}$

Step 3.5.2.4.1.7

Rewrite $432$ as $12^{2} \cdot 3$ .

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Step 3.5.2.4.1.7.1

Factor $144$ out of $432$ .

$b = \frac{- 12 \pm i \cdot \sqrt{144 (3)}}{2 \cdot 16}$

Step 3.5.2.4.1.7.2

Rewrite $144$ as $12^{2}$ .

$b = \frac{- 12 \pm i \cdot \sqrt{12^{2} \cdot 3}}{2 \cdot 16}$

Step 3.5.2.4.1.8

Pull terms out from under the radical.

$b = \frac{- 12 \pm i \cdot (12 \sqrt{3})}{2 \cdot 16}$

Step 3.5.2.4.1.9

Move $12$ to the left of $i$ .

$b = \frac{- 12 \pm 12 i \sqrt{3}}{2 \cdot 16}$

Step 3.5.2.4.2

Multiply $2$ by $16$ .

$b = \frac{- 12 \pm 12 i \sqrt{3}}{32}$

Step 3.5.2.4.3

Simplify $\frac{- 12 \pm 12 i \sqrt{3}}{32}$ .

$b = \frac{- 3 \pm 3 i \sqrt{3}}{8}$

Step 3.5.2.5

The final answer is the combination of both solutions.

$b = - \frac{3 - 3 i \sqrt{3}}{8}, - \frac{3 + 3 i \sqrt{3}}{8}$

Step 3.6

The final solution is all the values that make $(b - \frac{3}{4}) (b^{2} + \frac{3 b}{4} + \frac{9}{16}) = 0$ true.

$b = \frac{3}{4}, - \frac{3 - 3 i \sqrt{3}}{8}, - \frac{3 + 3 i \sqrt{3}}{8}$