Solve for B log base B of 64=3

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

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 $b^{y} = x$ .

$B^{3} = 64$

Solve for $B$ .

Subtract $64$ from both sides of the equation.

$B^{3} - 64 = 0$

Factor the left side of the equation.

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

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

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 $b = 4$ .

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

Simplify.

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

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

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

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

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

$B - 4 = 0$

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

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

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

$B - 4 = 0$

Add $4$ to both sides of the equation.

$B = 4$

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

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

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

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $16$ .

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

Subtract $64$ from $16$ .

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

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

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

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

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

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

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

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

Factor $16$ out of $48$ .

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

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

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

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

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

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

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

The final answer is the combination of both solutions.

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

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

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

Solve for B log base B of 64=3

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