Algebra Examples

$\log_{8} (2) + \log_{8} (4 x^{2}) = 1$

Step 1

Simplify the left side.

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Step 1.1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

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

Step 1.2

Multiply $4$ by $2$ .

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

Step 2

Rewrite $\log_{8} (8 x^{2}) = 1$ 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 $b^{y} = x$ .

$8^{1} = 8 x^{2}$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $8 x^{2} = 8^{1}$ .

$8 x^{2} = 8$

Step 3.2

Divide each term in $8 x^{2} = 8^{1}$ by $8$ and simplify.

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Step 3.2.1

Divide each term in $8 x^{2} = 8^{1}$ by $8$ .

$\frac{8 x^{2}}{8} = \frac{8^{1}}{8}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $8$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{8 x^{2}}{8} = \frac{8^{1}}{8}$

Step 3.2.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{8^{1}}{8}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Cancel the common factor of $8^{1}$ and $8$ .

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Step 3.2.3.1.1

Factor $8$ out of $8^{1}$ .

$x^{2} = \frac{8 \cdot 1}{8}$

Step 3.2.3.1.2

Cancel the common factors.

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Step 3.2.3.1.2.1

Factor $8$ out of $8$ .

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

Step 3.2.3.1.2.2

Cancel the common factor.

$x^{2} = \frac{8 \cdot 1}{8 \cdot 1}$

Step 3.2.3.1.2.3

Rewrite the expression.

$x^{2} = \frac{1}{1}$

Step 3.2.3.1.2.4

Rewrite the expression.

$x^{2} = 1$

Step 3.3

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

$x = \pm \sqrt{1}$

Step 3.4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 3.5

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

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Step 3.5.1

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

$x = 1$

Step 3.5.2

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

$x = - 1$

Step 3.5.3

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

$x = 1, - 1$