Problem 34
Question
Use logarithms to solve each problem. How long will it take an investment of $$\$ 5000$$ to triple if the investment earns interest at the rate of \(8 \% /\) year compounded daily?
Step-by-Step Solution
Verified Answer
It will take approximately \(13.94\) years for an investment of $5000 to triple at a \(8\%\) interest rate compounded daily.
1Step 1: Write down the compound interest formula
We are given the compound interest formula:
\[A = P(1 + r/n)^{nt}\]
2Step 2: Substitute the known values
We know that the initial investment (P) is $5,000, the annual interest rate (r) is 8%, and the interest is compounded daily (n = 365). We also know that the final value (A) is three times the initial investment.
So, our equation becomes:
\[3(5000) = 5000(1 + \frac{0.08}{365})^{365t}\]
3Step 3: Simplify the equation
Divide both sides of the equation by $5,000 to remove the initial investment (P) and simplify as needed:
\[3 = (1 + \frac{0.08}{365})^{365t}\]
4Step 4: Use logarithms to isolate t
To solve for t, we can apply a logarithm to both sides of the equation. Since we are dealing with an exponential equation, it is helpful to use the natural logarithm (ln). Apply ln to both sides of our simplified equation:
\[\ln{3} = \ln{(1 + \frac{0.08}{365})^{365t}}\]
Now, use the properties of logarithms to bring down the exponent (365t) in front of the ln:
\[\ln{3} = 365t\ln{(1 + \frac{0.08}{365})}\]
5Step 5: Solve for t
Now, we need to isolate t by dividing both sides of the equation by \(365\ln{(1 + \frac{0.08}{365})}\):
\[t = \frac{\ln{3}}{365\ln{(1 + \frac{0.08}{365})}}\]
6Step 6: Calculate the value of t
Finally, plug the values into a calculator to find the approximate value of t:
\[t \approx \frac{\ln{3}}{365\ln{(1 + \frac{0.08}{365})}} \approx 13.94\]
It will take approximately 13.94 years for the initial investment of $5,000 to triple at an interest rate of 8% compounded daily.
Key Concepts
Compound Interest FormulaExponential EquationsSolving for Time with Logarithms
Compound Interest Formula
The compound interest formula is a crucial tool for calculating how investments grow over time. In mathematical terms, the formula is represented as:\[A = P(1 + \frac{r}{n})^{nt}\]Where:
- \(A\) is the future value of the investment, or the amount of money accumulated after the stated time, including interest.
- \(P\) is the principal investment amount, which is the initial amount of money invested.
- \(r\) is the annual interest rate expressed in decimal form. For example, if the interest rate is 8%, \(r\) would be 0.08.
- \(n\) is the number of times interest is compounded per year. In the case of daily compounding, \(n\) would be 365.
- \(t\) is the time in years for which the money is invested or borrowed.
Exponential Equations
Exponential equations involve variables as exponents. They often emerge in growth scenarios such as population growth, radioactive decay, and financial investments. In these equations, the variable is positioned in the exponent, making them uniquely challenging compared to linear or quadratic equations.
In our context, for the investment to triple, the final value \(A\) will be three times the initial investment \(P\), turning our compound interest equation into:\[3(5000) = 5000(1 + \frac{0.08}{365})^{365t}\]After simplifying, it becomes:\[3 = (1 + \frac{0.08}{365})^{365t}\]Exponential equations like this one are solved by isolating the variable on one side of the equation. In such cases, logarithms are employed to "bring down" the exponents to the level of the other operations.
In our context, for the investment to triple, the final value \(A\) will be three times the initial investment \(P\), turning our compound interest equation into:\[3(5000) = 5000(1 + \frac{0.08}{365})^{365t}\]After simplifying, it becomes:\[3 = (1 + \frac{0.08}{365})^{365t}\]Exponential equations like this one are solved by isolating the variable on one side of the equation. In such cases, logarithms are employed to "bring down" the exponents to the level of the other operations.
Solving for Time with Logarithms
Logarithms are essential for solving exponential equations where the variable is in the exponent. They help by allowing us to isolate the time variable \(t\) in the compound interest context.
Starting with the simplified exponential equation:\[3 = (1 + \frac{0.08}{365})^{365t}\]we apply a logarithm to both sides. Using a natural logarithm (\(\ln\)), as it's particularly effective for these types of calculations:\[\ln{3} = \ln{(1 + \frac{0.08}{365})^{365t}}\]This allows us to use the property \(\ln{(x^y)} = y\ln{x}\), bringing the exponent \(365t\) to the front:\[\ln{3} = 365t\ln{(1 + \frac{0.08}{365})}\]Now, we solve for \(t\) by dividing both sides by \(365\ln{(1 + \frac{0.08}{365})}\):\[t = \frac{\ln{3}}{365\ln{(1 + \frac{0.08}{365})}}\]Calculating this, we find out that \(t\) is approximately 13.94 years. Thus, it will take just under 14 years for the investment to triple under the given conditions.
Starting with the simplified exponential equation:\[3 = (1 + \frac{0.08}{365})^{365t}\]we apply a logarithm to both sides. Using a natural logarithm (\(\ln\)), as it's particularly effective for these types of calculations:\[\ln{3} = \ln{(1 + \frac{0.08}{365})^{365t}}\]This allows us to use the property \(\ln{(x^y)} = y\ln{x}\), bringing the exponent \(365t\) to the front:\[\ln{3} = 365t\ln{(1 + \frac{0.08}{365})}\]Now, we solve for \(t\) by dividing both sides by \(365\ln{(1 + \frac{0.08}{365})}\):\[t = \frac{\ln{3}}{365\ln{(1 + \frac{0.08}{365})}}\]Calculating this, we find out that \(t\) is approximately 13.94 years. Thus, it will take just under 14 years for the investment to triple under the given conditions.
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