Precalculus Examples

$\ln (x^{2} - x + 2) = \ln (4)$

Step 1

Set the argument in $\ln (x^{2} - x + 2)$ greater than $0$ to find where the expression is defined.

$x^{2} - x + 2 > 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Convert the inequality to an equation.

$x^{2} - x + 2 = 0$

Step 2.2

Use the quadratic formula to find the solutions.

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

Step 2.3

Substitute the values $a = 1$ , $b = - 1$ , and $c = 2$ into the quadratic formula and solve for $x$ .

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

Step 2.4

Simplify.

Tap for more steps...

Step 2.4.1

Simplify the numerator.

Tap for more steps...

Step 2.4.1.1

Raise $- 1$ to the power of $2$ .

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

Step 2.4.1.2

Multiply $- 4 \cdot 1 \cdot 2$ .

Tap for more steps...

Step 2.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.4.1.2.2

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8}}{2 \cdot 1}$

Step 2.4.1.3

Subtract $8$ from $1$ .

$x = \frac{1 \pm \sqrt{- 7}}{2 \cdot 1}$

Step 2.4.1.4

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 7}}{2 \cdot 1}$

Step 2.4.1.5

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{7}}{2 \cdot 1}$

Step 2.4.1.6

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

$x = \frac{1 \pm i \sqrt{7}}{2 \cdot 1}$

Step 2.4.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{7}}{2}$

Step 2.5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Tap for more steps...

Step 2.5.1

Simplify the numerator.

Tap for more steps...

Step 2.5.1.1

Raise $- 1$ to the power of $2$ .

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

Step 2.5.1.2

Multiply $- 4 \cdot 1 \cdot 2$ .

Tap for more steps...

Step 2.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.5.1.2.2

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8}}{2 \cdot 1}$

Step 2.5.1.3

Subtract $8$ from $1$ .

$x = \frac{1 \pm \sqrt{- 7}}{2 \cdot 1}$

Step 2.5.1.4

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 7}}{2 \cdot 1}$

Step 2.5.1.5

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{7}}{2 \cdot 1}$

Step 2.5.1.6

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

$x = \frac{1 \pm i \sqrt{7}}{2 \cdot 1}$

Step 2.5.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{7}}{2}$

Step 2.5.3

Change the $\pm$ to $+$ .

$x = \frac{1 + i \sqrt{7}}{2}$

Step 2.6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Tap for more steps...

Step 2.6.1

Simplify the numerator.

Tap for more steps...

Step 2.6.1.1

Raise $- 1$ to the power of $2$ .

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

Step 2.6.1.2

Multiply $- 4 \cdot 1 \cdot 2$ .

Tap for more steps...

Step 2.6.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.6.1.2.2

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8}}{2 \cdot 1}$

Step 2.6.1.3

Subtract $8$ from $1$ .

$x = \frac{1 \pm \sqrt{- 7}}{2 \cdot 1}$

Step 2.6.1.4

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 7}}{2 \cdot 1}$

Step 2.6.1.5

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{7}}{2 \cdot 1}$

Step 2.6.1.6

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

$x = \frac{1 \pm i \sqrt{7}}{2 \cdot 1}$

Step 2.6.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{7}}{2}$

Step 2.6.3

Change the $\pm$ to $-$ .

$x = \frac{1 - i \sqrt{7}}{2}$

Step 2.7

Identify the leading coefficient.

Tap for more steps...

Step 2.7.1

The leading term in a polynomial is the term with the highest degree.

$x^{2}$

Step 2.7.2

The leading coefficient in a polynomial is the coefficient of the leading term.

$1$

Step 2.8

Since there are no real x-intercepts and the leading coefficient is positive, the parabola opens up and $x^{2} - x + 2$ is always greater than $0$ .

All real numbers

Step 3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4