Basic Math Examples

A n = {(- \frac{4}{n})}^{n - 1}

$A n = {(- \frac{4}{n})}^{n - 1}$

Step 1

Since

n

$n$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

{(- \frac{4}{n})}^{n - 1} = A n

${(- \frac{4}{n})}^{n - 1} = A n$

Step 2

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

ln ({(- \frac{4}{n})}^{n - 1}) = ln (A n)

$\ln ({(- \frac{4}{n})}^{n - 1}) = \ln (A n)$

Step 3

Expand

ln ({(- \frac{4}{n})}^{n - 1})

$\ln ({(- \frac{4}{n})}^{n - 1})$ by moving

n - 1

$n - 1$ outside the logarithm.

(n - 1) ln (- \frac{4}{n}) = ln (A n)

$(n - 1) \ln (- \frac{4}{n}) = \ln (A n)$

Step 4

Simplify the left side.

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Step 4.1

Simplify

(n - 1) ln (- \frac{4}{n})

$(n - 1) \ln (- \frac{4}{n})$ .

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Step 4.1.1

Apply the distributive property.

n ln (- \frac{4}{n}) - 1 ln (- \frac{4}{n}) = ln (A n)

$n \ln (- \frac{4}{n}) - 1 \ln (- \frac{4}{n}) = \ln (A n)$

Step 4.1.2

Rewrite

- 1 ln (- \frac{4}{n})

$- 1 \ln (- \frac{4}{n})$ as

- ln (- \frac{4}{n})

$- \ln (- \frac{4}{n})$ .

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (A n)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (A n)$

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (A n)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (A n)$

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (A n)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (A n)$

Step 5

Move all the terms containing a logarithm to the left side of the equation.

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) - ln (A n) = 0

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (A n) = 0$

Step 6

To solve for

n

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

e^{ln (A n)} = e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}

$e^{\ln (A n)} = e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}$

Step 7

Rewrite

ln (A n) = n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$\ln (A n) = n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ 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^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})} = A n

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = A n$

Step 8

Solve for

n

$n$ .

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Step 8.1

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

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}) = ln (A n)

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}) = \ln (A n)$

Step 8.2

Expand the left side.

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Step 8.2.1

Expand

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})})

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) ln (e) = ln (A n)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (A n)$

Step 8.2.2

The natural logarithm of

e

$e$ is

1

$1$ .

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) \cdot 1 = ln (A n)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (A n)$

Step 8.2.3

Multiply

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

1

$1$ .

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (A n)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (A n)$

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (A n)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (A n)$

Step 8.3

Move all the terms containing a logarithm to the left side of the equation.

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) - ln (A n) = 0

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (A n) = 0$

Step 8.4

To solve for

n

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

e^{ln (A n)} = e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}

$e^{\ln (A n)} = e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}$

Step 8.5

Rewrite

ln (A n) = n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$\ln (A n) = n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ 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^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})} = A n

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = A n$

Step 8.6

Solve for

n

$n$ .

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Step 8.6.1

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

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}) = ln (A n)

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}) = \ln (A n)$

Step 8.6.2

Expand the left side.

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Step 8.6.2.1

Expand

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})})

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) ln (e) = ln (A n)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (A n)$

Step 8.6.2.2

The natural logarithm of

e

$e$ is

1

$1$ .

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) \cdot 1 = ln (A n)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (A n)$

Step 8.6.2.3

Multiply

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

1

$1$ .

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (A n)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (A n)$

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (A n)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (A n)$

Step 8.6.3

Move all the terms containing a logarithm to the left side of the equation.

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) - ln (A n) = 0

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (A n) = 0$

Step 8.6.4

To solve for

n

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

e^{ln (A n)} = e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}

$e^{\ln (A n)} = e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}$

Step 8.6.5

Rewrite

ln (A n) = n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$\ln (A n) = n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ 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^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})} = A n

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = A n$

Step 8.6.6

Solve for

n

$n$ .

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Step 8.6.6.1

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

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}) = ln (A n)

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}) = \ln (A n)$

Step 8.6.6.2

Expand the left side.

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Step 8.6.6.2.1

Expand

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})})

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) ln (e) = ln (A n)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (A n)$

Step 8.6.6.2.2

The natural logarithm of

e

$e$ is

1

$1$ .

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) \cdot 1 = ln (A n)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (A n)$

Step 8.6.6.2.3

Multiply

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

1

$1$ .

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (A n)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (A n)$

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (A n)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (A n)$

Step 8.6.6.3

Move all the terms containing a logarithm to the left side of the equation.

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) - ln (A n) = 0

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (A n) = 0$

Step 8.6.6.4

To solve for

n

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

e^{ln (A n)} = e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}

$e^{\ln (A n)} = e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}$

Step 8.6.6.5

Rewrite

ln (A n) = n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$\ln (A n) = n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ 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^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})} = A n

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = A n$

Step 8.6.6.6

Solve for

n

$n$ .

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Step 8.6.6.6.1

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

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}) = ln (A n)

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}) = \ln (A n)$

Step 8.6.6.6.2

Expand the left side.

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Step 8.6.6.6.2.1

Expand

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})})

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) ln (e) = ln (A n)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (A n)$

Step 8.6.6.6.2.2

The natural logarithm of

e

$e$ is

1

$1$ .

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) \cdot 1 = ln (A n)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (A n)$

Step 8.6.6.6.2.3

Multiply

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

1

$1$ .

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (A n)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (A n)$

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (A n)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (A n)$

Step 8.6.6.6.3

Subtract

ln (A n)

$\ln (A n)$ from both sides of the equation.

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) - ln (A n) = 0

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (A n) = 0$

Step 8.6.6.6.4

Reorder

A

$A$ and

n

$n$ .

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) - ln (n A) = 0

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (n A) = 0$

Step 8.6.6.6.5

To solve for

n

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

e^{ln (n A)} = e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}

$e^{\ln (n A)} = e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}$

Step 8.6.6.6.6

Rewrite

ln (n A) = n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$\ln (n A) = n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ 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^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})} = n A

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A$

Step 8.6.6.6.7

Solve for

n

$n$ .

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Step 8.6.6.6.7.1

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

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}) = ln (n A)

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}) = \ln (n A)$

Step 8.6.6.6.7.2

Expand the left side.

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Step 8.6.6.6.7.2.1

Expand

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})})

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) ln (e) = ln (n A)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (n A)$

Step 8.6.6.6.7.2.2

The natural logarithm of

e

$e$ is

1

$1$ .

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) \cdot 1 = ln (n A)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (n A)$

Step 8.6.6.6.7.2.3

Multiply

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

1

$1$ .

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (n A)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (n A)$

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (n A)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (n A)$

Step 8.6.6.6.7.3

Move all the terms containing a logarithm to the left side of the equation.

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) - ln (n A) = 0

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (n A) = 0$

Step 8.6.6.6.7.4

To solve for

n

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

e^{ln (n A)} = e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}

$e^{\ln (n A)} = e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}$

Step 8.6.6.6.7.5

Rewrite

ln (n A) = n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$\ln (n A) = n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ 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^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})} = n A

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A$

Step 8.6.6.6.7.6

Solve for

n

$n$ .

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Step 8.6.6.6.7.6.1

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

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}) = ln (n A)

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}) = \ln (n A)$

Step 8.6.6.6.7.6.2

Expand the left side.

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Step 8.6.6.6.7.6.2.1

Expand

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})})

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) ln (e) = ln (n A)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (n A)$

Step 8.6.6.6.7.6.2.2

The natural logarithm of

e

$e$ is

1

$1$ .

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) \cdot 1 = ln (n A)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (n A)$

Step 8.6.6.6.7.6.2.3

Multiply

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

1

$1$ .

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (n A)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (n A)$

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (n A)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (n A)$

Step 8.6.6.6.7.6.3

Move all the terms containing a logarithm to the left side of the equation.

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) - ln (n A) = 0

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (n A) = 0$

Step 8.6.6.6.7.6.4

To solve for

n

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

e^{ln (n A)} = e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}

$e^{\ln (n A)} = e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}$

Step 8.6.6.6.7.6.5

Rewrite

ln (n A) = n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$\ln (n A) = n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ 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^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})} = n A

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A$

Step 8.6.6.6.7.6.6

Solve for

n

$n$ .

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Step 8.6.6.6.7.6.6.1

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

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}) = ln (n A)

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}) = \ln (n A)$

Step 8.6.6.6.7.6.6.2

Expand the left side.

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Step 8.6.6.6.7.6.6.2.1

Expand

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})})

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) ln (e) = ln (n A)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (n A)$

Step 8.6.6.6.7.6.6.2.2

The natural logarithm of

e

$e$ is

1

$1$ .

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) \cdot 1 = ln (n A)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (n A)$

Step 8.6.6.6.7.6.6.2.3

Multiply

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

1

$1$ .

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (n A)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (n A)$

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (n A)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (n A)$

Step 8.6.6.6.7.6.6.3

Move all the terms containing a logarithm to the left side of the equation.

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) - ln (n A) = 0

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (n A) = 0$

Step 8.6.6.6.7.6.6.4

To solve for

n

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

e^{ln (n A)} = e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}

$e^{\ln (n A)} = e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}$

Step 8.6.6.6.7.6.6.5

Rewrite

ln (n A) = n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$\ln (n A) = n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ 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^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})} = n A

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A$

Step 8.6.6.6.7.6.6.6

Solve for

n

$n$ .

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Step 8.6.6.6.7.6.6.6.1

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

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}) = ln (n A)

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.2

Expand the left side.

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Step 8.6.6.6.7.6.6.6.2.1

Expand

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})})

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) ln (e) = ln (n A)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.2.2

The natural logarithm of

e

$e$ is

1

$1$ .

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) \cdot 1 = ln (n A)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (n A)$

Step 8.6.6.6.7.6.6.6.2.3

Multiply

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

1

$1$ .

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (n A)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (n A)$

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (n A)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.3

Move all the terms containing a logarithm to the left side of the equation.

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) - ln (n A) = 0

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (n A) = 0$

Step 8.6.6.6.7.6.6.6.4

To solve for

n

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

e^{ln (n A)} = e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}

$e^{\ln (n A)} = e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}$

Step 8.6.6.6.7.6.6.6.5

Rewrite

ln (n A) = n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$\ln (n A) = n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ 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^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})} = n A

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A$

Step 8.6.6.6.7.6.6.6.6

Solve for

n

$n$ .

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Step 8.6.6.6.7.6.6.6.6.1

Subtract

n A

$n A$ from both sides of the equation.

e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})} - n A = 0

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} - n A = 0$

Step 8.6.6.6.7.6.6.6.6.2

Move all the terms containing a logarithm to the left side of the equation.

e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})} = n A + 0

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A + 0$

Step 8.6.6.6.7.6.6.6.6.3

Add

n A

$n A$ and

0

$0$ .

e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})} = n A

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A$

Step 8.6.6.6.7.6.6.6.6.4

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

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})}) = ln (n A)

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.5

Expand the left side.

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Step 8.6.6.6.7.6.6.6.6.5.1

Expand

ln (e^{n ln (- \frac{4}{n}) - ln (- \frac{4}{n})})

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) ln (e) = ln (n A)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.5.2

The natural logarithm of

e

$e$ is

1

$1$ .

(n ln (- \frac{4}{n}) - ln (- \frac{4}{n})) \cdot 1 = ln (n A)

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.5.3

Multiply

n ln (- \frac{4}{n}) - ln (- \frac{4}{n})

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

1

$1$ .

n ln (- \frac{4}{n}) - ln (- \frac{4}{n}) = ln (n A)

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.6

Subtract

$\ln (n A)$ from both sides of the equation.

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (n A) = 0$

Step 8.6.6.6.7.6.6.6.6.7

To solve for

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

$e^{\ln (n A)} = e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}$

Step 8.6.6.6.7.6.6.6.6.8

Rewrite

$\ln (n A) = n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ 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

$b^{y} = x$ .

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A$

Step 8.6.6.6.7.6.6.6.6.9

Solve for

$n$ .

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Step 8.6.6.6.7.6.6.6.6.9.1

Subtract

$n A$ from both sides of the equation.

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} - n A = 0$

Step 8.6.6.6.7.6.6.6.6.9.2

Move all the terms containing a logarithm to the left side of the equation.

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A + 0$

Step 8.6.6.6.7.6.6.6.6.9.3

Add

$n A$ and

$0$ .

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A$

Step 8.6.6.6.7.6.6.6.6.9.4

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

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.9.5

Expand the left side.

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Step 8.6.6.6.7.6.6.6.6.9.5.1

Expand

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.9.5.2

The natural logarithm of

$e$ is

$1$ .

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.9.5.3

Multiply

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

$1$ .

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.9.6

Subtract

$\ln (n A)$ from both sides of the equation.

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (n A) = 0$

Step 8.6.6.6.7.6.6.6.6.9.7

To solve for

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

$e^{\ln (n A)} = e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}$

Step 8.6.6.6.7.6.6.6.6.9.8

Rewrite

$\ln (n A) = n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ 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

$b^{y} = x$ .

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A$

Step 8.6.6.6.7.6.6.6.6.9.9

Solve for

$n$ .

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Step 8.6.6.6.7.6.6.6.6.9.9.1

Subtract

$n A$ from both sides of the equation.

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} - n A = 0$

Step 8.6.6.6.7.6.6.6.6.9.9.2

Move all the terms containing a logarithm to the left side of the equation.

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A + 0$

Step 8.6.6.6.7.6.6.6.6.9.9.3

Add

$n A$ and

$0$ .

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A$

Step 8.6.6.6.7.6.6.6.6.9.9.4

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

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.9.9.5

Expand the left side.

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Step 8.6.6.6.7.6.6.6.6.9.9.5.1

Expand

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.9.9.5.2

The natural logarithm of

$e$ is

$1$ .

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.9.9.5.3

Multiply

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

$1$ .

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.9.9.6

Subtract

$\ln (n A)$ from both sides of the equation.

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) - \ln (n A) = 0$

Step 8.6.6.6.7.6.6.6.6.9.9.7

To solve for

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

$e^{\ln (n A)} = e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}$

Step 8.6.6.6.7.6.6.6.6.9.9.8

Rewrite

$\ln (n A) = n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ 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

$b^{y} = x$ .

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A$

Step 8.6.6.6.7.6.6.6.6.9.9.9

Solve for

$n$ .

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Step 8.6.6.6.7.6.6.6.6.9.9.9.1

Subtract

$n A$ from both sides of the equation.

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} - n A = 0$

Step 8.6.6.6.7.6.6.6.6.9.9.9.2

Move all the terms containing a logarithm to the left side of the equation.

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A + 0$

Step 8.6.6.6.7.6.6.6.6.9.9.9.3

Add

$n A$ and

$0$ .

$e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})} = n A$

Step 8.6.6.6.7.6.6.6.6.9.9.9.4

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

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})}) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.9.9.9.5

Expand the left side.

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Step 8.6.6.6.7.6.6.6.6.9.9.9.5.1

Expand

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.9.9.9.5.2

The natural logarithm of

$e$ is

$1$ .

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (n A)$

Step 8.6.6.6.7.6.6.6.6.9.9.9.5.3

Multiply

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

$1$ .

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (n A)$

Step 8.6.6.6.8

Expand the left side.

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Step 8.6.6.6.8.1

Expand

$\ln (e^{n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})})$ by moving

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ outside the logarithm.

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \ln (e) = \ln (A n)$

Step 8.6.6.6.8.2

The natural logarithm of

$e$ is

$1$ .

$(n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})) \cdot 1 = \ln (A n)$

Step 8.6.6.6.8.3

Multiply

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n})$ by

$1$ .

$n \ln (- \frac{4}{n}) - \ln (- \frac{4}{n}) = \ln (A n)$

Step 9

Expand

$\ln ({(- \frac{4}{n})}^{n - 1})$ by moving

$n - 1$ outside the logarithm.

$(n - 1) \ln (- \frac{4}{n}) = \ln (A n)$