Pre-Algebra Examples

$2^{2 x} - 2^{x - 1} - 2^{2} + 2 < 0$

Step 1

Rewrite $2^{x - 1}$ as $2^{x} \cdot 2^{- 1}$ .

$2^{2 x} - (2^{x} \cdot 2^{- 1}) - 2^{2} + 2 = 0$

Step 2

Rewrite $2^{2 x}$ as exponentiation.

${(2^{x})}^{2} - (2^{x} \cdot 2^{- 1}) - 2^{2} + 2 = 0$

Step 3

Remove parentheses.

${(2^{x})}^{2} - 2^{x} \cdot 2^{- 1} - 2^{2} + 2 = 0$

Step 4

Substitute $u$ for $2^{x}$ .

$u^{2} - u \cdot 2^{- 1} - 2^{2} + 2 = 0$

Step 5

Simplify each term.

Tap for more steps...

Step 5.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$u^{2} - u \cdot \frac{1}{2} - 2^{2} + 2 = 0$

Step 5.2

Combine $\frac{1}{2}$ and $u$ .

$u^{2} - \frac{u}{2} - 2^{2} + 2 = 0$

Step 5.3

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

$u^{2} - \frac{u}{2} - 1 \cdot 4 + 2 = 0$

Step 5.4

Multiply $- 1$ by $4$ .

$u^{2} - \frac{u}{2} - 4 + 2 = 0$

Step 6

Add $- 4$ and $2$ .

$u^{2} - \frac{u}{2} - 2 = 0$

Step 7

Solve for $u$ .

Tap for more steps...

Step 7.1

Multiply through by the least common denominator $2$ , then simplify.

Tap for more steps...

Step 7.1.1

Apply the distributive property.

$2 u^{2} + 2 (- \frac{u}{2}) + 2 \cdot - 2 = 0$

Step 7.1.2

Simplify.

Tap for more steps...

Step 7.1.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 7.1.2.1.1

Move the leading negative in $- \frac{u}{2}$ into the numerator.

$2 u^{2} + 2 (\frac{- u}{2}) + 2 \cdot - 2 = 0$

Step 7.1.2.1.2

Cancel the common factor.

$2 u^{2} + 2 (\frac{- u}{2}) + 2 \cdot - 2 = 0$

Step 7.1.2.1.3

Rewrite the expression.

$2 u^{2} - u + 2 \cdot - 2 = 0$

Step 7.1.2.2

Multiply $2$ by $- 2$ .

$2 u^{2} - u - 4 = 0$

Step 7.2

Use the quadratic formula to find the solutions.

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

Step 7.3

Substitute the values $a = 2$ , $b = - 1$ , and $c = - 4$ into the quadratic formula and solve for $u$ .

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

Step 7.4

Simplify.

Tap for more steps...

Step 7.4.1

Simplify the numerator.

Tap for more steps...

Step 7.4.1.1

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

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

Step 7.4.1.2

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

Tap for more steps...

Step 7.4.1.2.1

Multiply $- 4$ by $2$ .

$u = \frac{1 \pm \sqrt{1 - 8 \cdot - 4}}{2 \cdot 2}$

Step 7.4.1.2.2

Multiply $- 8$ by $- 4$ .

$u = \frac{1 \pm \sqrt{1 + 32}}{2 \cdot 2}$

Step 7.4.1.3

Add $1$ and $32$ .

$u = \frac{1 \pm \sqrt{33}}{2 \cdot 2}$

Step 7.4.2

Multiply $2$ by $2$ .

$u = \frac{1 \pm \sqrt{33}}{4}$

Step 7.5

The final answer is the combination of both solutions.

$u = \frac{1 + \sqrt{33}}{4}, \frac{1 - \sqrt{33}}{4}$

Step 8

Substitute $\frac{1 + \sqrt{33}}{4}$ for $u$ in $u = 2^{x}$ .

$\frac{1 + \sqrt{33}}{4} = 2^{x}$

Step 9

Solve $\frac{1 + \sqrt{33}}{4} = 2^{x}$ .

Tap for more steps...

Step 9.1

Rewrite the equation as $2^{x} = \frac{1 + \sqrt{33}}{4}$ .

$2^{x} = \frac{1 + \sqrt{33}}{4}$

Step 9.2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (2^{x}) = \ln (\frac{1 + \sqrt{33}}{4})$

Step 9.3

Expand $\ln (2^{x})$ by moving $x$ outside the logarithm.

$x \ln (2) = \ln (\frac{1 + \sqrt{33}}{4})$

Step 9.4

Divide each term in $x \ln (2) = \ln (\frac{1 + \sqrt{33}}{4})$ by $\ln (2)$ and simplify.

Tap for more steps...

Step 9.4.1

Divide each term in $x \ln (2) = \ln (\frac{1 + \sqrt{33}}{4})$ by $\ln (2)$ .

$\frac{x \ln (2)}{\ln (2)} = \frac{\ln (\frac{1 + \sqrt{33}}{4})}{\ln (2)}$

Step 9.4.2

Simplify the left side.

Tap for more steps...

Step 9.4.2.1

Cancel the common factor of $\ln (2)$ .

Tap for more steps...

Step 9.4.2.1.1

Cancel the common factor.

$\frac{x \ln (2)}{\ln (2)} = \frac{\ln (\frac{1 + \sqrt{33}}{4})}{\ln (2)}$

Step 9.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (\frac{1 + \sqrt{33}}{4})}{\ln (2)}$

Step 10

Substitute $\frac{1 - \sqrt{33}}{4}$ for $u$ in $u = 2^{x}$ .

$\frac{1 - \sqrt{33}}{4} = 2^{x}$

Step 11

Solve $\frac{1 - \sqrt{33}}{4} = 2^{x}$ .

Tap for more steps...

Step 11.1

Rewrite the equation as $2^{x} = \frac{1 - \sqrt{33}}{4}$ .

$2^{x} = \frac{1 - \sqrt{33}}{4}$

Step 11.2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (2^{x}) = \ln (\frac{1 - \sqrt{33}}{4})$

Step 11.3

The equation cannot be solved because $\ln (\frac{1 - \sqrt{33}}{4})$ is undefined.

Undefined

Step 11.4

There is no solution for $2^{x} = \frac{1 - \sqrt{33}}{4}$

No solution

Step 12

List the solutions that makes the equation true.

$x = \frac{\ln (\frac{1 + \sqrt{33}}{4})}{\ln (2)}$

Step 13

The solution consists of all of the true intervals.

$x < \frac{\ln (\frac{1 + \sqrt{33}}{4})}{\ln (2)}$

Step 14

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{\ln (\frac{1 + \sqrt{33}}{4})}{\ln (2)}$

Interval Notation:

$(- \infty, \frac{\ln (\frac{1 + \sqrt{33}}{4})}{\ln (2)})$

Step 15