How long will it take money to double if it is invested
(a) at 8% compounded daily?
(b) at 7.1 % compounded continuously?
Given information:
Amount is doubled.
It means if we invest $x, then it becomes $2x in some years.
Formula for part A:
We use compound interest formula.
A= P(1+r/n)^(nt)
Where “A” is amount after ‘t’ years. P is initial investment. ‘r’ is rate of interest in decimal. ‘n’ is number of times compounded in a year.
r = 8%= 8/100 =0.08
n=365
Plug in the known values into the formula to get value for ‘t’.
2x =x(1+0.08/365)^(365t)
Divide both sides by x.
2= (1+0.08/365)^(365t)
Simplify the numbers inside the parentheses,
2= (1.000219178)^(365t)
Now, take log on both sides to bring exponent 365t to front of parentheses.
log 2 = log (1.000219178)^365t
Using power rule of logarithm we can rewrite it as,
log 2 = 365t( log 1.000219178)
Simplify using calculator.
0.3010299 =0.034739738t
Divide both sides by 0.034739738
8.665=t
Round it to nearest years.
So, t =9 years.
Answer to part a:
It will take about ___9___ years at 8% compounded daily
Part B:
(b) at 7.1 % compounded continuously?
r=7.1% =7.1/100= 0.071
It says compounded continuously, so we have specific formula for this.
A = P e^(rt)
We know P is $x, A is $2x. r is 0.071 and we need to find ‘t’.
Plug in the known values into the formula.
2x = x e^(0.071t)
Divide both sides by x.
2= e^(0.071t)
Take natural log on both sides.
ln(2)= ln e^(0.071t)
Simplify both sides,
0.69314718 = 0.071t
Divide both sides by 0.071
t =9.76 years
Answer to part b:
It will take about __10___ years at 7.1% compounded continuously.