Question

the formula $a = pe^{rt}$ describes the accumulated value, $a$, of a sum of money, $p$, the principal, after $t$ years at annual percentage rate $r$ (in decimal form) compounded continuously. complete the table for a savings account subject to continuous compounding.

amount invested	annual interest rate	accumulated amount	time $t$ in years

(do not round until the final answer. then round to one decimal place as needed.)

Explanation:

Step1: Identify given values

We know the formula for continuous compounding is \( A = Pe^{rt} \). Here, \( P = 5500 \), \( r = 10\% = 0.1 \), and \( A = 2\times5500 = 11000 \) (since we need to double the amount invested).

Step2: Substitute values into the formula

Substitute \( A = 11000 \), \( P = 5500 \), and \( r = 0.1 \) into \( A = Pe^{rt} \):

$$ 11000 = 5500e^{0.1t} $$

Step3: Solve for \( t \)

First, divide both sides by 5500:

$$ \frac{11000}{5500}=e^{0.1t} $$

Simplify the left side: \( 2 = e^{0.1t} \)
Take the natural logarithm of both sides:

$$ \ln(2)=\ln(e^{0.1t}) $$

Using the property \( \ln(e^x)=x \), we get:

$$ \ln(2)=0.1t $$

Then solve for \( t \) by dividing both sides by 0.1:

$$ t=\frac{\ln(2)}{0.1} $$

Calculate \( \ln(2)\approx0.6931 \), so:

$$ t=\frac{0.6931}{0.1}=6.931 $$

Round to one decimal place: \( t\approx6.9 \)

Answer:

\( 6.9 \)