Precalculus Examples

$\ln (x^{6}) = 12$

Step 1

Write in exponential form.

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Step 1.1

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, $b = e$ , $x = x^{6}$ , and $y = 12$ .

$b = e$

$x = x^{6}$

$y = 12$

Step 1.2

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

$e^{12} = x^{6}$

Step 2

Solve for $x$ .

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Step 2.1

Rewrite the equation as $x^{6} = e^{12}$ .

$x^{6} = e^{12}$

Step 2.2

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

$x = \pm \sqrt[6]{e^{12}}$

Step 2.3

Simplify $\pm \sqrt[6]{e^{12}}$ .

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Step 2.3.1

Rewrite $e^{12}$ as ${(e^{2})}^{6}$ .

$x = \pm \sqrt[6]{{(e^{2})}^{6}}$

Step 2.3.2

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

$x = \pm e^{2}$

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.

$x = e^{2}$

Step 2.4.2

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

$x = - e^{2}$

Step 2.4.3

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

$x = e^{2}, - e^{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = e^{2}, - e^{2}$

Decimal Form:

$x = 7.38905609 \dots, - 7.38905609 \dots$