Problem 72 - A Graphical Approach to Precalculus with Limits

Question

Solve each formula for the indicated variable. $$A=\frac{P i}{1-(1+i)^{-n}}, \text { for } n$$

Step-by-Step Solution

Answer

\( n = \frac{-\ln\left(1 - \frac{Pi}{A}\right)}{\ln(1+i)} \).

Step 1: Rearrange the Formula

Our given formula is:\[ A = \frac{Pi}{1-(1+i)^{-n}} \]We want to solve it for \( n \). To start, rearrange the formula to isolate the fraction on one side.

Step 2: Express in Terms of the Denominator

Multiply both sides by the denominator to eliminate the fraction:\[ A \cdot \left(1-(1+i)^{-n}\right) = Pi \]

Step 3: Solve for Denominator Expression

Divide both sides by \( A \) to isolate the expression involving \( n \) on one side:\[ 1-(1+i)^{-n} = \frac{Pi}{A} \]

Step 4: Isolate Power Term

Subtract the result from 1 to isolate the negative power term:\[ (1+i)^{-n} = 1 - \frac{Pi}{A} \]

Step 5: Use Logarithms to Solve for n

Take the natural logarithm of both sides to solve for \( n \):\[ -n \ln(1+i) = \ln\left(1 - \frac{Pi}{A}\right) \]

Step 6: Final Expression for n

Solve for \( n \) by dividing by \(-\ln(1+i)\):\[ n = \frac{-\ln\left(1 - \frac{Pi}{A}\right)}{\ln(1+i)} \]

Key Concepts

Algebraic Manipulation

Algebraic manipulation is the process of rearranging equations to isolate a variable or simplify an expression. It involves using arithmetic and algebraic operations like addition, subtraction, multiplication, division, and exponentiation.

In our exercise, we start with the equation:

• First, the goal is to rearrange this equation to find the variable of interest, which in this case is \( n \), present in the denominator of a fraction. To do this, you might need to apply several algebraic operations, always working step by step to maintain equality.
• Multiply both sides by the denominator to eliminate the fraction, which is a crucial step to simplify the expression and make it workable.
• Then, isolate the terms involving \( n \) by performing division and subtraction where necessary.

This method allows for a systematic approach to finding a variable within a complex equation. The key to mastering algebraic manipulation lies in practicing these fundamental operations and being thorough with each step.

Logarithms

Logarithms are an essential concept in algebra and are used to solve equations where the variable is an exponent. A logarithm answers the question: "To what power must the base be raised, to yield a certain number?"

In the solution provided:

• The natural logarithm \( \ln \) is used to bring down the variable \( n \) from the exponent. The equation contained the term \( (1+i)^{-n} \), which can be simplified using logarithms.
• By taking the logarithm of both sides of the equation, the exponent \( -n \) is brought down, turning the equation into a linear form which can be more easily solved.
• Once the variable is isolated, further algebraic steps can proceed easily to derive a solution for \( n \).

Using logarithms in this way is a powerful tool, especially when dealing with equations that involve exponentials, as it transforms a potentially complex equation into a simpler, linear form.

Mathematical Formulas

A mathematical formula is essentially a concise way of expressing information symbolically. These formulas help describe relationships and outputs from certain inputs.

For the formula provided:

• We are concerned with solving for \( n \) within the formula \( A = \frac{Pi}{1-(1+i)^{-n}} \). This formula is used in fields such as finance for calculating payments, annuities, and interest.
• Solving this specifically for the variable \( n \) shows us how the period of time relates to other financial parameters like principal, interest rate, and total amount.
• Formulas are invaluable in structuring these relationships and calculations, enabling quick computation without the need to derive each relationship from scratch.

Understanding a formula involves knowing what each part stands for and how altering a single variable affects the entire expression. Practicing with different formulas enhances your foundational knowledge of mathematical relationships.