Precalculus Examples

${(2^{2 x} \cdot 2^{2 x})}^{x - 1} = 8$

Step 1

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

$\ln ({(2^{2 x} \cdot 2^{2 x})}^{x - 1}) = \ln (8)$

Step 2

Expand the left side.

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Step 2.1

Expand $\ln ({(2^{2 x} \cdot 2^{2 x})}^{x - 1})$ by moving $x - 1$ outside the logarithm.

$(x - 1) \ln (2^{2 x} \cdot 2^{2 x}) = \ln (8)$

Step 2.2

Multiply $2^{2 x}$ by $2^{2 x}$ by adding the exponents.

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Step 2.2.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$(x - 1) \ln (2^{2 x + 2 x}) = \ln (8)$

Step 2.2.2

Add $2 x$ and $2 x$ .

$(x - 1) \ln (2^{4 x}) = \ln (8)$

Step 2.3

Expand $\ln (2^{4 x})$ by moving $4 x$ outside the logarithm.

$(x - 1) (4 x \ln (2)) = \ln (8)$

Step 3

Simplify the left side.

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Step 3.1

Simplify $(x - 1) (4 x \ln (2))$ .

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Step 3.1.1

Simplify by multiplying through.

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Step 3.1.1.1

Apply the distributive property.

$x (4 x \ln (2)) - 1 (4 x \ln (2)) = \ln (8)$

Step 3.1.1.2

Simplify the expression.

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Step 3.1.1.2.1

Rewrite using the commutative property of multiplication.

$4 \ln (2) x \cdot x - 1 (4 x \ln (2)) = \ln (8)$

Step 3.1.1.2.2

Multiply $4$ by $- 1$ .

$4 \ln (2) x \cdot x - 4 (x \ln (2)) = \ln (8)$

Step 3.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 3.1.2.1

Move $x$ .

$4 \ln (2) (x \cdot x) - 4 (x \ln (2)) = \ln (8)$

Step 3.1.2.2

Multiply $x$ by $x$ .

$4 \ln (2) x^{2} - 4 (x \ln (2)) = \ln (8)$

$4 \ln (2) x^{2} - 4 x \ln (2) = \ln (8)$

Step 3.1.3

Reorder factors in $4 \ln (2) x^{2} - 4 x \ln (2)$ .

$4 x^{2} \ln (2) - 4 x \ln (2) = \ln (8)$

Step 4

Move all the terms containing a logarithm to the left side of the equation.

$4 x^{2} \ln (2) - 4 x \ln (2) - \ln (8) = 0$

Step 5

Use the quadratic formula to find the solutions.

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

Step 6

Substitute the values $a = 4 \ln (2)$ , $b = - 4 \ln (2)$ , and $c = - \ln (8)$ into the quadratic formula and solve for $x$ .

$\frac{4 \ln (2) \pm \sqrt{{(- 4 \ln (2))}^{2} - 4 \cdot (4 \ln (2) \cdot (- \ln (8)))}}{2 (4 \ln (2))}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Add parentheses.

$x = \frac{4 \ln (2) \pm \sqrt{{(- 4 \ln (2))}^{2} - 4 \cdot (4 (\ln (2) \cdot (- \ln (8))))}}{2 \cdot (4 \ln (2))}$

Step 7.1.2

Let $u = 4 (\ln (2) \cdot (- \ln (8)))$ . Substitute $u$ for all occurrences of $4 (\ln (2) \cdot (- \ln (8)))$ .

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Step 7.1.2.1

Apply the product rule to $- 4 \ln (2)$ .

$x = \frac{4 \ln (2) \pm \sqrt{{(- 4)}^{2} \ln^{2} (2) - 4 \cdot u}}{2 \cdot (4 \ln (2))}$

Step 7.1.2.2

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

$x = \frac{4 \ln (2) \pm \sqrt{16 \ln^{2} (2) - 4 u}}{2 \cdot (4 \ln (2))}$

Step 7.1.3

Factor $4$ out of $16 \ln^{2} (2) - 4 u$ .

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Step 7.1.3.1

Factor $4$ out of $16 \ln^{2} (2)$ .

$x = \frac{4 \ln (2) \pm \sqrt{4 (4 \ln^{2} (2)) - 4 u}}{2 \cdot (4 \ln (2))}$

Step 7.1.3.2

Factor $4$ out of $- 4 u$ .

$x = \frac{4 \ln (2) \pm \sqrt{4 (4 \ln^{2} (2)) + 4 (- u)}}{2 \cdot (4 \ln (2))}$

Step 7.1.3.3

Factor $4$ out of $4 (4 \ln^{2} (2)) + 4 (- u)$ .

$x = \frac{4 \ln (2) \pm \sqrt{4 (4 \ln^{2} (2) - u)}}{2 \cdot (4 \ln (2))}$

Step 7.1.4

Replace all occurrences of $u$ with $4 (\ln (2) \cdot (- \ln (8)))$ .

$x = \frac{4 \ln (2) \pm \sqrt{4 (4 \ln^{2} (2) - (4 (\ln (2) \cdot (- \ln (8)))))}}{2 \cdot (4 \ln (2))}$

Step 7.1.5

Simplify each term.

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Step 7.1.5.1

Rewrite using the commutative property of multiplication.

$x = \frac{4 \ln (2) \pm \sqrt{4 (4 \ln^{2} (2) - (4 (- \ln (2) \ln (8))))}}{2 \cdot (4 \ln (2))}$

Step 7.1.5.2

Multiply $- 1$ by $4$ .

$x = \frac{4 \ln (2) \pm \sqrt{4 (4 \ln^{2} (2) - (- 4 (\ln (2) \ln (8))))}}{2 \cdot (4 \ln (2))}$

Step 7.1.5.3

Multiply $- 4$ by $- 1$ .

$x = \frac{4 \ln (2) \pm \sqrt{4 (4 \ln^{2} (2) + 4 \ln (2) \ln (8))}}{2 \cdot (4 \ln (2))}$

Step 7.1.6

Factor $4 \ln (2)$ out of $4 \ln^{2} (2) + 4 \ln (2) \ln (8)$ .

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Step 7.1.6.1

Factor $4 \ln (2)$ out of $4 \ln^{2} (2)$ .

$x = \frac{4 \ln (2) \pm \sqrt{4 (4 \ln (2) \ln (2) + 4 \ln (2) \ln (8))}}{2 \cdot (4 \ln (2))}$

Step 7.1.6.2

Factor $4 \ln (2)$ out of $4 \ln (2) \ln (8)$ .

$x = \frac{4 \ln (2) \pm \sqrt{4 (4 \ln (2) \ln (2) + 4 \ln (2) \ln (8))}}{2 \cdot (4 \ln (2))}$

Step 7.1.6.3

Factor $4 \ln (2)$ out of $4 \ln (2) \ln (2) + 4 \ln (2) \ln (8)$ .

$x = \frac{4 \ln (2) \pm \sqrt{4 (4 \ln (2) (\ln (2) + \ln (8)))}}{2 \cdot (4 \ln (2))}$

$x = \frac{4 \ln (2) \pm \sqrt{4 \cdot (4 \ln (2) (\ln (2) + \ln (8)))}}{2 \cdot (4 \ln (2))}$

Step 7.1.7

Multiply $4$ by $4$ .

$x = \frac{4 \ln (2) \pm \sqrt{16 \ln (2) (\ln (2) + \ln (8))}}{2 \cdot (4 \ln (2))}$

Step 7.1.8

Rewrite $16 \ln (2) (\ln (2) + \ln (8))$ as ${(2^{2})}^{2} (\ln (2) (\ln (2) + \ln (8)))$ .

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Step 7.1.8.1

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

$x = \frac{4 \ln (2) \pm \sqrt{4^{2} \ln (2) (\ln (2) + \ln (8))}}{2 \cdot (4 \ln (2))}$

Step 7.1.8.2

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

$x = \frac{4 \ln (2) \pm \sqrt{{(2^{2})}^{2} \ln (2) (\ln (2) + \ln (8))}}{2 \cdot (4 \ln (2))}$

Step 7.1.8.3

Add parentheses.

$x = \frac{4 \ln (2) \pm \sqrt{{(2^{2})}^{2} (\ln (2) (\ln (2) + \ln (8)))}}{2 \cdot (4 \ln (2))}$

Step 7.1.9

Pull terms out from under the radical.

$x = \frac{4 \ln (2) \pm 2^{2} \sqrt{\ln (2) (\ln (2) + \ln (8))}}{2 \cdot (4 \ln (2))}$

Step 7.1.10

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

$x = \frac{4 \ln (2) \pm 4 \sqrt{\ln (2) (\ln (2) + \ln (8))}}{2 \cdot (4 \ln (2))}$

Step 7.2

Multiply $2$ by $4$ .

$x = \frac{4 \ln (2) \pm 4 \sqrt{\ln (2) (\ln (2) + \ln (8))}}{8 \ln (2)}$

Step 7.3

Simplify $\frac{4 \ln (2) \pm 4 \sqrt{\ln (2) (\ln (2) + \ln (8))}}{8 \ln (2)}$ .

$x = \frac{\ln (2) \pm \sqrt{\ln (2) (\ln (2) + \ln (8))}}{2 \ln (2)}$

Step 8

The final answer is the combination of both solutions.

$x = \frac{\ln (2) + \sqrt{\ln (2) (\ln (2) + \ln (8))}}{2 \ln (2)}, \frac{\ln (2) - \sqrt{\ln (2) (\ln (2) + \ln (8))}}{2 \ln (2)}$

Step 9

The result can be shown in multiple forms.

Exact Form:

$x = \frac{\ln (2) + \sqrt{\ln (2) (\ln (2) + \ln (8))}}{2 \ln (2)}, \frac{\ln (2) - \sqrt{\ln (2) (\ln (2) + \ln (8))}}{2 \ln (2)}$

Decimal Form:

$x = 1.5, - 0.5$