Question 148856
Suppose that $5000 is invested at interest rate k, compounded continuously, and grows to $6954.84 in 6 years.
:
The continuous interest equation: {{{A = P(e^(rt))}}}



 a) Find the interest rate. (find k)
{{{6954.84 = 5000(e^(6k))}}}
:
{{{e^(6k)}}} = {{{6954.84/5000}}}
:
{{{e^(6k)}}} = 1.390968
Find the nat log of both sides
{{{ln(e^(6k))}}} = ln(1.390968)
ln of e is 1
6k = .330
k = {{{.330/6}}}
k = .055 or 5.5% interest
:
Check solution on a calc: enter 5000(e^(.055*6)) = 6954.840
: 
:
b) write the exponential growth function.
{{{A = P(e^(rt))}}}
:
:
 c) Find the balance after 10years.
on Calc enter: 5000(e^(.055*10)) = 8666.27