Calculus Examples

e^{ln (x^{2})} - 16 = 0

$e^{\ln (x^{2})} - 16 = 0$

Step 1

Add

16

$16$ to both sides of the equation.

e^{ln (x^{2})} = 16

$e^{\ln (x^{2})} = 16$

Step 2

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

ln (e^{ln (x^{2})}) = ln (16)

$\ln (e^{\ln (x^{2})}) = \ln (16)$

Step 3

Expand the left side.

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Step 3.1

Expand

ln (e^{ln (x^{2})})

$\ln (e^{\ln (x^{2})})$ by moving

ln (x^{2})

$\ln (x^{2})$ outside the logarithm.

ln (x^{2}) ln (e) = ln (16)

$\ln (x^{2}) \ln (e) = \ln (16)$

Step 3.2

The natural logarithm of

e

$e$ is

1

$1$ .

ln (x^{2}) \cdot 1 = ln (16)

$\ln (x^{2}) \cdot 1 = \ln (16)$

Step 3.3

Multiply

ln (x^{2})

$\ln (x^{2})$ by

1

$1$ .

ln (x^{2}) = ln (16)

$\ln (x^{2}) = \ln (16)$

ln (x^{2}) = ln (16)

$\ln (x^{2}) = \ln (16)$

Step 4

To solve for

x

$x$ , rewrite the equation using properties of logarithms.

e^{ln (x^{2})} = e^{ln (16)}

$e^{\ln (x^{2})} = e^{\ln (16)}$

Step 5

Rewrite

ln (x^{2}) = ln (16)

$\ln (x^{2}) = \ln (16)$ 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

$b^{y} = x$ .

e^{ln (16)} = x^{2}

$e^{\ln (16)} = x^{2}$

Step 6

Solve for

x

$x$ .

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Step 6.1

Rewrite the equation as

x^{2} = e^{ln (16)}

$x^{2} = e^{\ln (16)}$ .

x^{2} = e^{ln (16)}

$x^{2} = e^{\ln (16)}$

Step 6.2

Exponentiation and log are inverse functions.

x^{2} = 16

$x^{2} = 16$

Step 6.3

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

x = \pm \sqrt{16}

$x = \pm \sqrt{16}$

Step 6.4

Simplify

\pm \sqrt{16}

$\pm \sqrt{16}$ .

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Step 6.4.1

Rewrite

16

$16$ as

4^{2}

$4^{2}$ .

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

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

Step 6.4.2

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

x = \pm 4

$x = \pm 4$

x = \pm 4

$x = \pm 4$

Step 6.5

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

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Step 6.5.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

x = 4

$x = 4$

Step 6.5.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

x = - 4

$x = - 4$

Step 6.5.3

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

x = 4, - 4

$x = 4, - 4$

x = 4, - 4

$x = 4, - 4$

x = 4, - 4

$x = 4, - 4$