Algebra Examples

$\log_{| x + 2 |} (64) = 3$

Step 1

Rewrite $\log_{| x + 2 |} (64) = 3$ 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 $b^{y} = x$ .

${| x + 2 |}^{3} = 64$

Step 2

Solve for $x$ .

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Step 2.1

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

$| x + 2 | = \sqrt[3]{64}$

Step 2.2

Simplify $\sqrt[3]{64}$ .

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Step 2.2.1

Rewrite $64$ as $4^{3}$ .

$| x + 2 | = \sqrt[3]{4^{3}}$

Step 2.2.2

Pull terms out from under the radical, assuming real numbers.

$| x + 2 | = 4$

Step 2.3

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x + 2 = \pm 4$

Step 2.4

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

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Step 2.4.1

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

$x + 2 = 4$

Step 2.4.2

Move all terms not containing $x$ to the right side of the equation.

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Step 2.4.2.1

Subtract $2$ from both sides of the equation.

$x = 4 - 2$

Step 2.4.2.2

Subtract $2$ from $4$ .

$x = 2$

Step 2.4.3

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

$x + 2 = - 4$

Step 2.4.4

Move all terms not containing $x$ to the right side of the equation.

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Step 2.4.4.1

Subtract $2$ from both sides of the equation.

$x = - 4 - 2$

Step 2.4.4.2

Subtract $2$ from $- 4$ .

$x = - 6$

Step 2.4.5

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

$x = 2, - 6$