Problem 17 - A Graphical Approach to College Algebra

Question

An amount \(A\) in a bank account after \(x\) years is given by \(A(x)=1000(1.025)^{x}\) (a) How much is in the account after 3 years? (b) How much is in the account after 10 years? (c) After how many years will there be about \(\$ 1900\) in the account?

Step-by-Step Solution

Answer

(a) $1077, (b) $1280.08, (c) about 27 years.

Step 1: Substitute 3 years into the formula

For part (a), we need to find \(A(3)\). Substitute \(x = 3\) into the formula: \(A(3) = 1000(1.025)^3\).

Step 2: Calculate the amount for 3 years

Compute the expression: \(A(3) = 1000(1.025)^3 = 1000 \times 1.077\). This gives \(A(3) = 1077\).

Step 3: Substitute 10 years into the formula

For part (b), we need to find \(A(10)\). Substitute \(x = 10\) into the formula: \(A(10) = 1000(1.025)^{10}\).

Step 4: Calculate the amount for 10 years

Compute the expression: \(A(10) = 1000(1.2800845)\). This gives \(A(10) \approx 1280.08\).

Step 5: Set up equation for \(\$1900\)

For part (c), solve for \(x\) when \(A(x) = 1900\). Set up the equation: \(1000(1.025)^{x} = 1900\).

Step 6: Solve for \(x\) using log

Divide both sides by 1000: \((1.025)^x = 1.9\). Take the logarithm of both sides: \(\log(1.025^x) = \log(1.9)\). This simplifies to \(x \cdot \log(1.025) = \log(1.9)\).

Step 7: Calculate \(x\)

Solve for \(x\): \(x = \frac{\log(1.9)}{\log(1.025)}\). Calculating gives \(x \approx 26.63\).

Key Concepts

Understanding Compound Interest

Compound interest is a powerful financial concept that suggests the interest earned on an account is added back to the principal, and future interest is then earned on this increased amount. This can result in exponential growth over time. Unlike simple interest, which only earns interest on the initial principal, compound interest takes previous interest into account, allowing the balance to grow at a faster rate. In the context of our original exercise, the formula \( A(x) = 1000(1.025)^x \) represents how the principal of \(1000 grows over several years with an annual interest rate of 2.5%.

To compute the future value, we substitute the number of years into the formula. For instance, \( A(3) \) indicates substituting 3 for \( x \), showing how much the original amount grows after 3 years which turns out to be \)1077 as we calculated. The formula efficiently predicts how much money we'll have in the future by applying the interest repeatedly over time.

Solving with Logarithmic Equations

Logarithmic equations are invaluable tools when deciphering the number of years needed for compound interest to grow a principal to a certain amount. Logarithms reverse the process of exponentiation, making them particularly useful for solving equations where the exponent is unknown.

In part (c) of our exercise, we determine how long it will take for \(1000 to grow to approximately \(\\)1900\). First, we set up the equation \( 1000(1.025)^x = 1900\) and simplify it to \((1.025)^x = 1.9\). To solve for \(x\), we use the logarithmic property: \( x = \frac{\log(1.9)}{\log(1.025)} \).

This approach uses the fact that the logarithm provides a way to "bring down" the exponent, making \( x \) easily solvable as \( x \approx 26.63 \). This means it would take a little over 26 years for the balance to grow from \(1000 to around \)1900 at the given interest rate.

Algebraic Problem Solving in Financial Contexts

Solving algebraic problems within financial contexts often involves setting up equations based on given formulas and inputs. The key is identifying what is needed and systematically applying algebraic techniques to find unknown values.

Our exercise provided a clear example of this process:

• First, for calculating the amount after a specific number of years, substitute the year into the exponential formula and solve.
• Next, for determining how long to achieve a particular goal amount, rearrange the equation and use logarithmic properties to isolate the unknown variable.

Algebraic problem solving is all about understanding the links between variables and constants, which allows us to translate real-world financial scenarios into solvable mathematical problems.

Mastering these skills not only equips you with tools to tackle textbook problems but also prepares you for real-life financial decision-making situations, like assessing loan terms, calculating investments, or planning savings.