Problem 102
Question
Find the time required for a \(\$ 1000\) investment to double at interest rate \(r\), compounded continuously.\(r=0.0875\)
Step-by-Step Solution
Verified Answer
To find the time it would take for a $1000 investment to double at an 0.0875 interest rate compounded continuously, calculate \( t = ln(2) / 0.0875 \). It yields approximately \(t \approx 7.93\) years.
1Step 1: Identify the Knowns
From the problem, we know that the initial principal amount (P) is $1000, the interest rate (r) is 0.0875, and the final amount (A) is double the initial investment, meaning A = $2000.
2Step 2: Applying the Formula
The continuous compound interest formula is \(A = P e^{rt}\). In this case, we are asked to solve for 't', so we first substitute the known values of P, A and r into the equation: \(2000 = 1000 * e^{0.0875t}\).
3Step 3: Solving for t
Divide both sides of equation by 1000, we get \(2 = e^{0.0875t}\). To solve for t, we take the natural log (ln) of both sides: ln(2) = \(0.0875t * ln(e)\), since \(ln(e) = 1\), the equation simplifies to \(0.0875t = ln(2)\). Thus, \( t = ln(2) / 0.0875 \).
4Step 4: Calculating the Final Value
By calculating this last equation, we can get the result for t.
Key Concepts
Exponential GrowthNatural LogarithmInvestment Doubling Time
Exponential Growth
In the world of finance and investment, exponential growth is a term used to describe an increase at a consistent rate over time. While linear growth adds the same amount each period, exponential growth involves a steady percentage increase, making it multiply rapidly as time goes on. It is often expressed using the formula:
\[ A = P e^{rt} \]
\[ A = P e^{rt} \]
- \(A\) is the amount of money accumulated after n years, including interest.
- \(P\) is the principal amount (the initial amount of money).
- \(r\) is the rate of interest per period.
- \(t\) is the time the money is invested for in years.
- \(e\) is the base of the natural logarithm, approximately equal to 2.71828.
Natural Logarithm
The natural logarithm, often denoted as \(ln\), is a special type of logarithm with the base \(e\), where \(e\) is approximately 2.71828. It is a commonly used method to solve equations involving exponential growth, especially those containing the \(e\) constant, like in continuous compounding. In our exercise, we applied the following steps:
In general, the natural logarithm function helps us unveil how certain processes grow or decay exponentially. It is ideal for calculating timeframes in scenarios involving continuous compounding interest, population growth, or decay in physics and biology.
- Starting with \(2 = e^{0.0875t}\), to find \(t\), we took the natural logarithm on both sides to get \(ln(2) = 0.0875t \,ln(e)\).
- Since \(ln(e) = 1\), it simplifies the equation to \(0.0875t = ln(2)\).
- We then isolated \(t\) by dividing both sides by 0.0875. This gives us \(t = ln(2) / 0.0875\).
In general, the natural logarithm function helps us unveil how certain processes grow or decay exponentially. It is ideal for calculating timeframes in scenarios involving continuous compounding interest, population growth, or decay in physics and biology.
Investment Doubling Time
Investment doubling time refers to the time required for an investment to grow to twice its original size. A classic way to estimate this is by using the Rule of 72.
However, for precise calculation, especially for continuous compounding as in the exercise, we rely on the formula:
\[ t = \frac{ln(2)}{r} \]
In our textbook example:
However, for precise calculation, especially for continuous compounding as in the exercise, we rely on the formula:
\[ t = \frac{ln(2)}{r} \]
In our textbook example:
- The investment is \(1000 and we want to know how long it takes to reach \)2000.
- Interest rate \(r\) is 0.0875 and employing the formula above, we substitute \(ln(2)\) divided by \(0.0875\) to get \(t\).
Other exercises in this chapter
Problem 101
Find the time required for a \(\$ 1000\) investment to double at interest rate \(r\), compounded continuously.\(r=0.0725\)
View solution Problem 101
Condense the expression to the logarithm of a single quantity.\(\ln x-\ln (x+2)-\ln (x-2)\)
View solution Problem 102
Condense the expression to the logarithm of a single quantity.\(\ln (x+1)+2 \ln (x-1)+3 \ln x\)
View solution Problem 103
The number \(V\) of varieties of suburban nondomesticated wildlife in a community is approximated by the model \(V=15 \cdot 10^{0.02 x}, \quad 0 \leq x \leq 36\
View solution