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.