Precalculus Examples

${\log_{x}}^{2} (x) - \log_{2} ({(x)}^{2}) = 15$

Step 1

Simplify each term.

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Step 1.1

Logarithm base $x$ of $x$ is $1$ .

$1^{2} - \log_{2} (x^{2}) = 15$

Step 1.2

One to any power is one.

$1 - \log_{2} (x^{2}) = 15$

Step 2

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

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Step 2.1

Subtract $1$ from both sides of the equation.

$- \log_{2} (x^{2}) = 15 - 1$

Step 2.2

Subtract $1$ from $15$ .

$- \log_{2} (x^{2}) = 14$

Step 3

Divide each term in $- \log_{2} (x^{2}) = 14$ by $- 1$ and simplify.

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Step 3.1

Divide each term in $- \log_{2} (x^{2}) = 14$ by $- 1$ .

$\frac{- \log_{2} (x^{2})}{- 1} = \frac{14}{- 1}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Dividing two negative values results in a positive value.

$\frac{\log_{2} (x^{2})}{1} = \frac{14}{- 1}$

Step 3.2.2

Divide $\log_{2} (x^{2})$ by $1$ .

$\log_{2} (x^{2}) = \frac{14}{- 1}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Divide $14$ by $- 1$ .

$\log_{2} (x^{2}) = - 14$

Step 4

Write in exponential form.

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Step 4.1

For logarithmic equations, $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ such that $x > 0$ , $b > 0$ , and $b \neq 1$ . In this case, $b = 2$ , $x = x^{2}$ , and $y = - 14$ .

$b = 2$

$x = x^{2}$

$y = - 14$

Step 4.2

Substitute the values of $b$ , $x$ , and $y$ into the equation $b^{y} = x$ .

$2^{- 14} = x^{2}$

Step 5

Solve for $x$ .

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Step 5.1

Rewrite the equation as $x^{2} = 2^{- 14}$ .

$x^{2} = 2^{- 14}$

Step 5.2

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

$x = \pm \sqrt{2^{- 14}}$

Step 5.3

Simplify $\pm \sqrt{2^{- 14}}$ .

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Step 5.3.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x = \pm \sqrt{\frac{1}{2^{14}}}$

Step 5.3.2

Raise $2$ to the power of $14$ .

$x = \pm \sqrt{\frac{1}{16384}}$

Step 5.3.3

Rewrite $\sqrt{\frac{1}{16384}}$ as $\frac{\sqrt{1}}{\sqrt{16384}}$ .

$x = \pm \frac{\sqrt{1}}{\sqrt{16384}}$

Step 5.3.4

Any root of $1$ is $1$ .

$x = \pm \frac{1}{\sqrt{16384}}$

Step 5.3.5

Simplify the denominator.

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Step 5.3.5.1

Rewrite $16384$ as $128^{2}$ .

$x = \pm \frac{1}{\sqrt{128^{2}}}$

Step 5.3.5.2

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

$x = \pm \frac{1}{128}$

Step 5.4

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

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Step 5.4.1

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

$x = \frac{1}{128}$

Step 5.4.2

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

$x = - \frac{1}{128}$

Step 5.4.3

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

$x = \frac{1}{128}, - \frac{1}{128}$

Step 6

Exclude the solutions that do not make ${\log_{x}}^{2} (x) - \log_{2} ({(x)}^{2}) = 15$ true.

$x = \frac{1}{128}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{128}$

Decimal Form:

$x = 0.0078125$