Problem 45
Question
(a) What annual rate of interest, compounded continuously for 100 years, would have multiplied Benjamin Franklin's original capital by \(90\) ? (b) In Benjamin Franklin's estimate that the original 1000 pounds would grow to \(131,000\) in 100 years, he was using an annual rate of 5\(\%\) and compounding once each year. What rate of interest per year when compounded continuously for 100 years would multiply the original amount by 131 ?
Step-by-Step Solution
Verified Answer
The annual interest rate for part (a), compounded continuously for 100 years, is given by \(\frac{1}{100} \ln(90)\). The rate of interest which, when compounded continuously for 100 years, would multiply the original amount by 131, is given by \(\frac{1}{100} \ln(131)\). The actual value of these expressions can be computed with a calculator to obtain numeric interest rates.
1Step 1: Computing the interest rate for part (a)
Start by rearranging the formula to solve for \(r\) : \[r = \frac{1}{t} \ln(\frac{A}{P})\]. Given that \(A = 90P\), \(P = 1\), and \(t = 100\), plug these values into the formula to find \(r\). After these substitutions, the calculation becomes \[r = \frac{1}{100} \ln(\frac{90}{1}).\] The calculated rate \(r\) for part (a) can be found by computing this expression.
2Step 2: Computing the interest rate for part (b)
The calculation for part (b) is similar to part (a). Simply replace \(A\) with \(131\) : \[r = \frac{1}{100} \ln(\frac{131}{1})\]. The calculated rate \(r\) for part (b) can be found by computing this expression.
Key Concepts
Continuously Compounded InterestLogarithmic FunctionsExponential Growth
Continuously Compounded Interest
When it comes to understanding interest calculations, continuous compounding can be a slightly intricate concept but very powerful. Unlike simple or annual compounding, continuously compounded interest calculates the interest on a principal sum continuously, hence, the interest is added at every possible moment in time.
This method utilizes an exponential growth model which means, with continuous compounding, the process of adding interest never truly stops.
It does so by employing the mathematical constant \(e\), approximately equal to 2.71828, in the formula:
This method utilizes an exponential growth model which means, with continuous compounding, the process of adding interest never truly stops.
It does so by employing the mathematical constant \(e\), approximately equal to 2.71828, in the formula:
- The formula used for calculating the future value under continuous compounding is: \[ A = Pe^{rt} \] where \(A\) is the amount of money accumulated after n years, including interest.
- \(P\) is the principal amount (the initial amount of money).
- \(r\) is the annual interest rate (expressed as a decimal).
- \(t\) is the time in years.
Logarithmic Functions
Logarithmic functions play an essential role in various fields of study, including finance, to unravel complexities in exponential growth models. To comprehend this, it's important to understand what a logarithm is.
At its core, a logarithm is the inverse operation to exponentiation. This means it undoes what exponentiation does. Simply put, if you have an equation like \(b^x = y\), then \(\log_b(y) = x\).
In contexts like continuously compounded interest, we often use the natural logarithm (denoted as \(\ln\)), which employs the base \(e\) (approximately 2.71828). Why do we use this? Well, whenever you have growth that changes at every moment, the natural logarithm provides a seamless way to solve for unknowns such as the interest rate \(r\):
At its core, a logarithm is the inverse operation to exponentiation. This means it undoes what exponentiation does. Simply put, if you have an equation like \(b^x = y\), then \(\log_b(y) = x\).
In contexts like continuously compounded interest, we often use the natural logarithm (denoted as \(\ln\)), which employs the base \(e\) (approximately 2.71828). Why do we use this? Well, whenever you have growth that changes at every moment, the natural logarithm provides a seamless way to solve for unknowns such as the interest rate \(r\):
- If we rearrange the formula for continuous compounding: \[ A = Pe^{rt} \] to solve for \(r\), we get: \[ r = \frac{1}{t} \ln\left(\frac{A}{P}\right) \]
- The \(\ln\) function helps us express the time it takes for investments to grow by a certain factor or determine the rate needed for a specific growth over time.
Exponential Growth
Exponential growth is a fundamental concept underpinning many natural and financial processes, most notably seen when interest compounds over time. Exponential growth establishes that as a quantity grows larger, its growth rate accelerates, often resulting in substantial increases over extended periods.
At the heart of exponential growth in financial contexts is the function involving the constant \(e\).
This growth arises from multiplying an initial quantity by a constant rate, which is adjustable in the formula:
In instances like Benjamin Franklin's investments, it illustrates how an initial sum can grow impressively if allowed to compound over long durations.
At the heart of exponential growth in financial contexts is the function involving the constant \(e\).
This growth arises from multiplying an initial quantity by a constant rate, which is adjustable in the formula:
- The general formula for exponential growth is expressed as: \[ A = P e^{rt} \] where \(e\) is the base of the natural logarithm.
- Each component of this equation reflects elements of growth:
- \(P\) represents the initial amount (or principal).
- \(e^{rt}\) depicts how multiplication by \(e\) at a rate \(r\), compounded over time \(t\), increases the initial amount.
In instances like Benjamin Franklin's investments, it illustrates how an initial sum can grow impressively if allowed to compound over long durations.
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