Precalculus Examples

{log}_{x} (16) = 4

$\log_{x} (16) = 4$

Step 1

Rewrite

{log}_{x} (16) = 4

$\log_{x} (16) = 4$ in exponential form using the definition of a logarithm. If

x

$x$ and

b

$b$ are positive real numbers and

b \neq 1

$b \neq 1$ , then

{log}_{b} (x) = y

$\log_{b} (x) = y$ is equivalent to

b^{y} = x

$b^{y} = x$ .

x^{4} = 16

$x^{4} = 16$

Step 2

Solve for

x

$x$ .

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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 = \pm \sqrt[4]{16}

$x = \pm \sqrt[4]{16}$

Step 2.2

Simplify

\pm \sqrt[4]{16}

$\pm \sqrt[4]{16}$ .

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Step 2.2.1

Rewrite

16

$16$ as

2^{4}

$2^{4}$ .

x = \pm \sqrt[4]{2^{4}}

$x = \pm \sqrt[4]{2^{4}}$

Step 2.2.2

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

x = \pm 2

$x = \pm 2$

x = \pm 2

$x = \pm 2$

Step 2.3

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

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Step 2.3.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

x = 2

$x = 2$

Step 2.3.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

x = - 2

$x = - 2$

Step 2.3.3

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

x = 2, - 2

$x = 2, - 2$

x = 2, - 2

$x = 2, - 2$

x = 2, - 2

$x = 2, - 2$

Step 3

Exclude the solutions that do not make

{log}_{x} (16) = 4

$\log_{x} (16) = 4$ true.

x = 2

$x = 2$

$\log_{x} (16) = 4$

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