If $4900 is invested in an account earning 7% annual interest compounded continuously, then the number of years that it takes for the amount to grow to $9800 is $\mathrm { n } = \frac { \ln 2 } { 0.07 }$ Find the number of years to the nearest tenth of a
Year.
A) 7.9 yr
B) 9.9 yr
C) 11.9 yr
D) 14.9 yr

Question

Quizplus · Accepted Answer

The formula for continuous compounding is $A = Pe^{rt}$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate, and $t$ is the time in years. To double the investment ($A/P = 2$), the formula simplifies to $2 = e^{0.07t}$. Taking the natural logarithm of both sides gives $\ln(2) = 0.07t$, which leads to $t = \frac{\ln(2)}{0.07}$. Calculating this gives approximately 9.9 years.

If $4900 Is Invested in an Account Earning 7% Annual n=ln⁡20.07\mathrm { n } = \frac { \ln 2 } { 0.07 }n=0.07ln2​

If $4900 Is Invested in an Account Earning 7% Annual $\mathrm { n } = \frac { \ln 2 } { 0.07 }$