QuestionJuly 29, 2025

How many years will it take for an initial investment of 20,000 to grow to 30,000 Assume a rate of interest of 10% compounded continuously. It will take about square years for the investment to grow to 30,000 (Round to two decimal places as needed.)

How many years will it take for an initial investment of 20,000 to grow to 30,000 Assume a rate of interest of 10% compounded continuously. It will take about square years for the investment to grow to 30,000 (Round to two decimal places as needed.)
How many years will it take for an initial investment of 20,000 to grow to 30,000 Assume a rate of interest of 10% 
compounded continuously.
It will take about square  years for the investment to grow to 30,000
(Round to two decimal places as needed.)

Solution
4.3(211 votes)

Answer

4.86 years Explanation 1. Use the Continuous Compounding Formula The formula for continuous compounding is **A = Pe^{rt}**, where A is the amount, P is the principal, r is the rate, and t is time. 2. Substitute Known Values Substitute A = 30000, P = 20000, and r = 0.10 into the formula: 30000 = 20000e^{0.10t}. 3. Solve for Time t Divide both sides by 20000: 1.5 = e^{0.10t}. Take the natural logarithm of both sides: \ln(1.5) = 0.10t. Solve for t: t = \frac{\ln(1.5)}{0.10}.

Explanation

1. Use the Continuous Compounding Formula<br /> The formula for continuous compounding is **$A = Pe^{rt}$**, where $A$ is the amount, $P$ is the principal, $r$ is the rate, and $t$ is time.<br />2. Substitute Known Values<br /> Substitute $A = 30000$, $P = 20000$, and $r = 0.10$ into the formula: $30000 = 20000e^{0.10t}$.<br />3. Solve for Time $t$<br /> Divide both sides by $20000$: $1.5 = e^{0.10t}$. Take the natural logarithm of both sides: $\ln(1.5) = 0.10t$. Solve for $t$: $t = \frac{\ln(1.5)}{0.10}$.
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