Basic Math Examples

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

Step 1

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

Expand $\ln ({(1 + r)}^{n})$ by moving $n$ outside the logarithm.

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

Step 3

The expanded equation is $\ln (p) + n \ln (1 + r) = \ln (f)$ .

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

Step 4

Simplify the left side.

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Step 4.1

Reorder $1$ and $r$ .

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

Step 4.2

Reorder $\ln (p)$ and $n \ln (r + 1)$ .

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

Step 5

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

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

Step 6

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 7

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

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

Step 8

Divide each term in $n \ln (r + 1) = - \ln (\frac{p}{f})$ by $n$ and simplify.

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Step 8.1

Divide each term in $n \ln (r + 1) = - \ln (\frac{p}{f})$ by $n$ .

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

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $n$ .

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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.

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