r = 7% Find the time necessary for $1000 to double when it is invested at a rate of r compounded (a) anually, (b) monthly, (c) daily, and (d) continuously

Formula for compounding n times per year A=P(1+r/n)^(nt)
Formula for compounding continuously A=Pe^(rt)
A=Final Amount
P=Initial Amount
r=rate of investment expressed as a decimal
n=number of compoundings per year
t= time in years
 
a) r=7%  n=1 (annually)
A=P(1+r/n)^(nt)
2000=1000(1+.07/1)^(1*t)
2=1.07^t
ln(2)=tln(1.07)
ln(2)/ln(1.07)=t
10.24=t
Final answer: 10.24 years
 
b) r=7% n=12 (monthly)
A=P(1+r/n)^(nt)
2000=1000(1+.07/12)^(12*t)
2=1.0058^(12t)
ln(2)=12tln(1.0058)
ln(2)/[12ln(1.0058)]=t
9.93=t
Final Answer: 9.93 years
 
c) r=7%  t=365 (daily)
A=P(1+r/n)^(nt)
2000=1000(1+.07/365)^(365*t)
2=(1.00019)^(365t)
ln(2)=365tln(1.00019)
ln(2)/[365ln(1.00019)]=t
9.90=t
Final answer: 9.90 years
 
d) r=7% compounded continously
A=Pe^(rt)
2000=1000e^(.07*t)
2=e^(.07t)
ln(2)=.07tlne
ln(2)/[.07lne]=t
9.90=t
Final answer: 9.90 years

Comments

Popular posts from this blog

Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 34

In “Fahrenheit 451,” what does Faber mean by “Those who don’t build must burn. It’s as old as history and juvenile delinquents”?

In what ways might RFID technology be used to serve customers better? What problems might arise? Do you think that the technology might be valuable when implanted in animals or people?