Precalculus Examples

$\log_{2} (8) + \log_{8} (x + 2) - \log_{4} (x^{4} + 4 x^{2} + 4)$

Step 1

Set the argument in $\log_{8} (x + 2)$ greater than $0$ to find where the expression is defined.

$x + 2 > 0$

Step 2

Subtract $2$ from both sides of the inequality.

$x > - 2$

Step 3

Set the argument in $\log_{4} (x^{4} + 4 x^{2} + 4)$ greater than $0$ to find where the expression is defined.

$x^{4} + 4 x^{2} + 4 > 0$

Step 4

Solve for $x$ .

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Step 4.1

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

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

$u = x^{2}$

Step 4.2

Factor using the perfect square rule.

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Step 4.2.1

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

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

Step 4.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 u = 2 \cdot u \cdot 2$

Step 4.2.3

Rewrite the polynomial.

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

Step 4.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 2$ .

${(u + 2)}^{2} = 0$

Step 4.3

Set the $u + 2$ equal to $0$ .

$u + 2 = 0$

Step 4.4

Subtract $2$ from both sides of the equation.

$u = - 2$

Step 4.5

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = - 2$

Step 4.6

Solve the equation for $x$ .

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Step 4.6.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 2}$

Step 4.6.2

Simplify $\pm \sqrt{- 2}$ .

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Step 4.6.2.1

Rewrite $- 2$ as $- 1 (2)$ .

$x = \pm \sqrt{- 1 (2)}$

Step 4.6.2.2

Rewrite $\sqrt{- 1 (2)}$ as $\sqrt{- 1} \cdot \sqrt{2}$ .

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

Step 4.6.2.3

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

$x = \pm i \sqrt{2}$

Step 4.6.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.6.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x = i \sqrt{2}$

Step 4.6.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i \sqrt{2}$

Step 4.6.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = i \sqrt{2}, - i \sqrt{2}$

Step 4.7

Identify the leading coefficient.

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Step 4.7.1

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

$x^{4}$

Step 4.7.2

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

$1$

Step 4.8

Since there are no real x-intercepts and the leading coefficient is positive, the parabola opens up and $x^{4} + 4 x^{2} + 4$ is always greater than $0$ .

All real numbers

Step 5

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- 2, \infty)$

Set-Builder Notation:

${x | x > - 2}$

Step 6