Solve for x log base 8 of 2+ log base 8 of 4x^2=1

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

Step 1

Simplify the left side.

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

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

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 $x$ .

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

$8 x^{2} = 8$

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

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

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

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

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

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

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

Simplify the right side.

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

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

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

Cancel the common factors.

Factor $8$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Rewrite the expression.

$x^{2} = 1$

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

$x = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

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

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

$x = 1$

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

$x = - 1$

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

$x = 1, - 1$

Solve for x log base 8 of 2+ log base 8 of 4x^2=1

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