Trigonometry Examples

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

Step 1

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

$b^{3} = 64$

Step 2

Solve for $b$ .

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Step 2.1

Subtract $64$ from both sides of the equation.

$b^{3} - 64 = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Rewrite $64$ as $4^{3}$ .

$b^{3} - 4^{3} = 0$

Step 2.2.2

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 $b = 4$ .

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

Step 2.2.3

Simplify.

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Step 2.2.3.1

Move $4$ to the left of $b$ .

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

Step 2.2.3.2

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

$(b - 4) (b^{2} + 4 b + 16) = 0$

Step 2.3

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

$b - 4 = 0$

$b^{2} + 4 b + 16 = 0$

Step 2.4

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

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Step 2.4.1

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

$b - 4 = 0$

Step 2.4.2

Add $4$ to both sides of the equation.

$b = 4$

Step 2.5

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

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Step 2.5.1

Set $b^{2} + 4 b + 16$ equal to $0$ .

$b^{2} + 4 b + 16 = 0$

Step 2.5.2

Solve $b^{2} + 4 b + 16 = 0$ for $b$ .

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Step 2.5.2.1

Use the quadratic formula to find the solutions.

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

Step 2.5.2.2

Substitute the values $a = 1$ , $b = 4$ , and $c = 16$ into the quadratic formula and solve for $b$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot 16)}}{2 \cdot 1}$

Step 2.5.2.3

Simplify.

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Step 2.5.2.3.1

Simplify the numerator.

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Step 2.5.2.3.1.1

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

$b = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Step 2.5.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 16$ .

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Step 2.5.2.3.1.2.1

Multiply $- 4$ by $1$ .

$b = \frac{- 4 \pm \sqrt{16 - 4 \cdot 16}}{2 \cdot 1}$

Step 2.5.2.3.1.2.2

Multiply $- 4$ by $16$ .

$b = \frac{- 4 \pm \sqrt{16 - 64}}{2 \cdot 1}$

Step 2.5.2.3.1.3

Subtract $64$ from $16$ .

$b = \frac{- 4 \pm \sqrt{- 48}}{2 \cdot 1}$

Step 2.5.2.3.1.4

Rewrite $- 48$ as $- 1 (48)$ .

$b = \frac{- 4 \pm \sqrt{- 1 \cdot 48}}{2 \cdot 1}$

Step 2.5.2.3.1.5

Rewrite $\sqrt{- 1 (48)}$ as $\sqrt{- 1} \cdot \sqrt{48}$ .

$b = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{48}}{2 \cdot 1}$

Step 2.5.2.3.1.6

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

$b = \frac{- 4 \pm i \cdot \sqrt{48}}{2 \cdot 1}$

Step 2.5.2.3.1.7

Rewrite $48$ as $4^{2} \cdot 3$ .

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Step 2.5.2.3.1.7.1

Factor $16$ out of $48$ .

$b = \frac{- 4 \pm i \cdot \sqrt{16 (3)}}{2 \cdot 1}$

Step 2.5.2.3.1.7.2

Rewrite $16$ as $4^{2}$ .

$b = \frac{- 4 \pm i \cdot \sqrt{4^{2} \cdot 3}}{2 \cdot 1}$

Step 2.5.2.3.1.8

Pull terms out from under the radical.

$b = \frac{- 4 \pm i \cdot (4 \sqrt{3})}{2 \cdot 1}$

Step 2.5.2.3.1.9

Move $4$ to the left of $i$ .

$b = \frac{- 4 \pm 4 i \sqrt{3}}{2 \cdot 1}$

Step 2.5.2.3.2

Multiply $2$ by $1$ .

$b = \frac{- 4 \pm 4 i \sqrt{3}}{2}$

Step 2.5.2.3.3

Simplify $\frac{- 4 \pm 4 i \sqrt{3}}{2}$ .

$b = - 2 \pm 2 i \sqrt{3}$

Step 2.5.2.4

The final answer is the combination of both solutions.

$b = - 2 + 2 i \sqrt{3}, - 2 - 2 i \sqrt{3}$

Step 2.6

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

$b = 4, - 2 + 2 i \sqrt{3}, - 2 - 2 i \sqrt{3}$