Solve for x log base x of 1000=3

$\log_{x} (1000) = 3$

Step 1

Rewrite $\log_{x} (1000) = 3$ 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 $b^{y} = x$ .

$x^{3} = 1000$

Step 2

Solve for $x$ .

Subtract $1000$ from both sides of the equation.

$x^{3} - 1000 = 0$

Factor the left side of the equation.

Rewrite $1000$ as $10^{3}$ .

$x^{3} - 10^{3} = 0$

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 $a = x$ and $b = 10$ .

$(x - 10) (x^{2} + x \cdot 10 + 10^{2}) = 0$

Simplify.

Move $10$ to the left of $x$ .

$(x - 10) (x^{2} + 10 x + 10^{2}) = 0$

Raise $10$ to the power of $2$ .

$(x - 10) (x^{2} + 10 x + 100) = 0$

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

$x - 10 = 0$

$x^{2} + 10 x + 100 = 0$

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

Set $x^{2} + 10 x + 100$ equal to $0$ and solve for $x$ .

Set $x^{2} + 10 x + 100$ equal to $0$ .

$x^{2} + 10 x + 100 = 0$

Solve $x^{2} + 10 x + 100 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = 10$ , and $c = 100$ into the quadratic formula and solve for $x$ .

$\frac{- 10 \pm \sqrt{10^{2} - 4 \cdot (1 \cdot 100)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $10$ to the power of $2$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot 1 \cdot 100}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 100$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 10 \pm \sqrt{100 - 4 \cdot 100}}{2 \cdot 1}$

Multiply $- 4$ by $100$ .

$x = \frac{- 10 \pm \sqrt{100 - 400}}{2 \cdot 1}$

Subtract $400$ from $100$ .

$x = \frac{- 10 \pm \sqrt{- 300}}{2 \cdot 1}$

Rewrite $- 300$ as $- 1 (300)$ .

$x = \frac{- 10 \pm \sqrt{- 1 \cdot 300}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (300)}$ as $\sqrt{- 1} \cdot \sqrt{300}$ .

$x = \frac{- 10 \pm \sqrt{- 1} \cdot \sqrt{300}}{2 \cdot 1}$

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

$x = \frac{- 10 \pm i \cdot \sqrt{300}}{2 \cdot 1}$

Rewrite $300$ as $10^{2} \cdot 3$ .

Factor $100$ out of $300$ .

$x = \frac{- 10 \pm i \cdot \sqrt{100 (3)}}{2 \cdot 1}$

Rewrite $100$ as $10^{2}$ .

$x = \frac{- 10 \pm i \cdot \sqrt{10^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 10 \pm i \cdot (10 \sqrt{3})}{2 \cdot 1}$

Move $10$ to the left of $i$ .

$x = \frac{- 10 \pm 10 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 10 \pm 10 i \sqrt{3}}{2}$

Simplify $\frac{- 10 \pm 10 i \sqrt{3}}{2}$ .

$x = - 5 \pm 5 i \sqrt{3}$

The final answer is the combination of both solutions.

$x = - 5 + 5 i \sqrt{3}, - 5 - 5 i \sqrt{3}$

The final solution is all the values that make $(x - 10) (x^{2} + 10 x + 100) = 0$ true.

$x = 10, - 5 + 5 i \sqrt{3}, - 5 - 5 i \sqrt{3}$

Solve for x log base x of 1000=3

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