Question 169194
Someone please help me with the following problem:
Suppose $5000 is invested at interest rate k, compounded continuously and grows to $6954.84 in 6 years.
:
The continuous interest formula: {{{P*e^(rt)}}} = A
Where:
P = initial amt (5000)
r = interest rate in decimal form (k in this problem)
t = time, in years (6yrs)
A = final amt (6954.84)
:
(a) Find the interest rate
{{{5000*e^(6k)}}} = 6954.84
{{{e^(6k)}}} = {{{6954.84/5000}}}
:
{{{e^(6k)}}} = 1.390968; divided both sides by 5000
:
6k*ln(e) = ln(1.390968); find the nat log of both sides
:
6k = .33; nat log of e is 1
:
k = {{{.33/6}}}
k = .055, 5.5% interest
:
Check on a calc: enter: 5000*e^(6*.055) = 6954.84
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(b) Find the exponential growth function
A = {{{5000*e^(.055t)}}}
;
:
(c) Find the balance after 10 years
On a calc enter {{{5000*e^((.055*10))}}} = 8666.27



(d) Find the doubling time
{{{5000*e^(.055t)}}} = 10000; we want to find t here
:
{{{e^(.055t)}}} = {{{10000/5000}}}
:
{{{e^(.055)}}} = 2; 
:
.055t*ln(e) = ln(2); find the nat log of both sides
:
.055t = .693; 
:
t = {{{.693/.055}}}
t = 12.6 yrs
:
:
Check on a calc: enter: 5000*e^(12.6*.055) = 9998.53 ~ 10000
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