Basic Math Examples

ln (f) = ln (p \cdot {(1 + r)}^{n})

$\ln (f) = \ln (p \cdot {(1 + r)}^{n})$

Step 1

Rewrite the equation as

ln (p \cdot {(1 + r)}^{n}) = ln (f)

$\ln (p \cdot {(1 + r)}^{n}) = \ln (f)$ .

ln (p \cdot {(1 + r)}^{n}) = ln (f)

$\ln (p \cdot {(1 + r)}^{n}) = \ln (f)$

Step 2

Expand

ln (p \cdot {(1 + r)}^{n})

$\ln (p \cdot {(1 + r)}^{n})$ .

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Step 2.1

Rewrite

ln (p \cdot {(1 + r)}^{n})

$\ln (p \cdot {(1 + r)}^{n})$ as

ln (p) + ln ({(1 + r)}^{n})

$\ln (p) + \ln ({(1 + r)}^{n})$ .

ln (p) + ln ({(1 + r)}^{n})

$\ln (p) + \ln ({(1 + r)}^{n})$

Step 2.2

Expand

ln ({(1 + r)}^{n})

$\ln ({(1 + r)}^{n})$ by moving

n

$n$ outside the logarithm.

ln (p) + n ln (1 + r)

$\ln (p) + n \ln (1 + r)$

ln (p) + n ln (1 + r)

$\ln (p) + n \ln (1 + r)$

Step 3

The expanded equation is

ln (p) + n ln (1 + r) = ln (f)

$\ln (p) + n \ln (1 + r) = \ln (f)$ .

ln (p) + n ln (1 + r) = ln (f)

$\ln (p) + n \ln (1 + r) = \ln (f)$

Step 4

Simplify the left side.

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Step 4.1

Reorder

1

$1$ and

r

$r$ .

ln (p) + n ln (r + 1) = ln (f)

$\ln (p) + n \ln (r + 1) = \ln (f)$

Step 4.2

Reorder

ln (p)

$\ln (p)$ and

n ln (r + 1)

$n \ln (r + 1)$ .

n ln (r + 1) + ln (p) = ln (f)

$n \ln (r + 1) + \ln (p) = \ln (f)$

n ln (r + 1) + ln (p) = ln (f)

$n \ln (r + 1) + \ln (p) = \ln (f)$

Step 5

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

n ln (r + 1) + ln (p) - ln (f) = 0

$n \ln (r + 1) + \ln (p) - \ln (f) = 0$

Step 6

Use the quotient property of logarithms,

{log}_{b} (x) - {log}_{b} (y) = {log}_{b} (\frac{x}{y})

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

n ln (r + 1) + ln (\frac{p}{f}) = 0

$n \ln (r + 1) + \ln (\frac{p}{f}) = 0$

Step 7

Subtract

ln (\frac{p}{f})

$\ln (\frac{p}{f})$ from both sides of the equation.

n ln (r + 1) = - ln (\frac{p}{f})

$n \ln (r + 1) = - \ln (\frac{p}{f})$

Step 8

Divide each term in

n ln (r + 1) = - ln (\frac{p}{f})

$n \ln (r + 1) = - \ln (\frac{p}{f})$ by

n

$n$ and simplify.

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Step 8.1

Divide each term in

n ln (r + 1) = - ln (\frac{p}{f})

$n \ln (r + 1) = - \ln (\frac{p}{f})$ by

n

$n$ .

\frac{n ln (r + 1)}{n} = \frac{- ln (\frac{p}{f})}{n}

$\frac{n \ln (r + 1)}{n} = \frac{- \ln (\frac{p}{f})}{n}$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of

n

$n$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{n \ln (r + 1)}{n} = \frac{- \ln (\frac{p}{f})}{n}$

Step 8.2.1.2

Divide

$\ln (r + 1)$ by

$1$ .

$\ln (r + 1) = \frac{- \ln (\frac{p}{f})}{n}$

Step 8.3

Simplify the right side.

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Step 8.3.1

Move the negative in front of the fraction.

$\ln (r + 1) = - \frac{\ln (\frac{p}{f})}{n}$

Step 9

To solve for

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

$e^{\ln (r + 1)} = e^{- \frac{\ln (\frac{p}{f})}{n}}$

Step 10

Rewrite

$\ln (r + 1) = - \frac{\ln (\frac{p}{f})}{n}$ 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

$b^{y} = x$ .

$e^{- \frac{\ln (\frac{p}{f})}{n}} = r + 1$

Step 11

Solve for

$r$ .

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Step 11.1

Rewrite the equation as

$r + 1 = e^{- \frac{\ln (\frac{p}{f})}{n}}$ .

$r + 1 = e^{- \frac{\ln (\frac{p}{f})}{n}}$

Step 11.2

Subtract

$1$ from both sides of the equation.

$r = e^{- \frac{\ln (\frac{p}{f})}{n}} - 1$