Question

a principal p, invested at 4 1/2% and compounded continuously, increases to an amount k times the original principal after t years, where t = ln(k)/0.045. (a) complete the table. (round your answers to one decimal place.) a table with columns t and k having rows for k = 1, 2, 4, 6, 8, 10, 12 with blank cells for t values interpret your result. the number of years required to multiply the original investment by an additional multiple of the original investment increases with k. however, the --select-- the value of k the fewer the years required to increase the value of the investment by an additional multiple of the original investment. (b) sketch a graph of the function.

Explanation:

Step1: Recall continuous - compounding formula

The formula for continuous - compounding is $A = Pe^{rt}$, where $A$ is the final amount, $P$ is the principal, $r$ is the annual interest rate, and $t$ is the number of years. We want to find $t$ when $A = KP$, so $KP=Pe^{rt}$. Canceling out $P$ gives $K = e^{rt}$. Solving for $t$: $t=\frac{\ln(K)}{r}$. Here $r = 0.045$.

Step2: Calculate $t$ for each value of $K$

When $K = 1$:
$t=\frac{\ln(1)}{0.045}=\frac{0}{0.045}=0.0$
When $K = 2$:
$t=\frac{\ln(2)}{0.045}\approx\frac{0.693147}{0.045}\approx15.4$
When $K = 4$:
$t=\frac{\ln(4)}{0.045}=\frac{2\ln(2)}{0.045}\approx\frac{2\times0.693147}{0.045}\approx30.8$
When $K = 6$:
$t=\frac{\ln(6)}{0.045}\approx\frac{1.791759}{0.045}\approx39.8$
When $K = 8$:
$t=\frac{\ln(8)}{0.045}=\frac{3\ln(2)}{0.045}\approx\frac{3\times0.693147}{0.045}\approx46.2$
When $K = 10$:
$t=\frac{\ln(10)}{0.045}\approx\frac{2.302585}{0.045}\approx51.2$
When $K = 12$:
$t=\frac{\ln(12)}{0.045}\approx\frac{2.484907}{0.045}\approx55.2$

Answer:

t	K
15.4	2
30.8	4
39.8	6
46.2	8
51.2	10
55.2	12