Continuous compound problem

Question

How long will it take for an investment of $1000.00 to grow to $6000.00 at an interest rate of 3% if interest is compounded

a) 3 times per year

b) Continuously

Answer 1

a. The formula for compound interest: A=P(1+r/n)^nt

Plugging in values

6000=1000(1+.03/3)^3t Now this can be solved. Divide both sides of the equation by 1000

6000/1000=1000(1+.03/3)^3t /1000 Simplify

6=(1.01)^3t Take the log of both sides of the equation

log(6)=3t(log(1.01)) Divide both sides by 3(log(1.01))

log(6)/[3(log(1.01))]=3t(log(1.01))/[3(log(1.01))] Simplify

60.0 = t

b. The formula for continuous compounding is A=Pe^rt

Plugging in values

6000=1000(e)^0.03t Now this can be solved. Divide both sides of the equation by 1000

6000/1000=1000(e)^0.03t/1000 Simplify

6=(e)^0.03t Take the ln of both sides

ln(6)=0.03t(ln(e)) Simplify

ln(6)=0.03t Divide both sides by 0.03

ln(6)/0/03=0.03t/0.03

59.7=t

Answer 2

For interest compounded for finite periods (monthly, quarterly, 3 times a year, etc), the equation is A = A₀(1 ± r/n)^n*t where A₀ is the initial amount, r is the annual rate in decimal form, n is the number of compounding periods, and t is the number of years. Additionally, you must decide whether to use a "+" or a "-" for the "±" by selecting "+" if the money is growing or "-" if the money is going down in value. In your example, we use a "+" because the money is growing. Also, we plug in the numbers this way:

6000 = 1000(1 + 0.03/3)^3t

To solve, first divide both sides by 1000:

6000/1000 = (1 + 0.03/3)^3t then take the log of both sides:

log(6) = log(1.01)^3t

Using the log property that an exponent can be brought out front of the log expression, this converts to:

log(6) = 3tlog(1.01) then dividing both sides by 3log(1.01) we get:

log(6)/[3log(1.01)] = t

Plugging this in a calculator we get t = 60.02 years

For interest compounded continuously, we use: A = A₀e^rt or, in your case:

6000 = 1000e^0.03t

6000/1000 = e^0.03t

ln(6) = ln(e^0.03t)

ln(6) = 0.03tln(e) and since ln(e) = 1 it becomes:

ln(6) = 0.03t

ln(6)/0.03 = t

t = 59.73 years

Answer 3

a) A = P(1+r/n)^nt where A is the final amount of money, P is the initial investment, r is the rate (need to convert to decimal), n is the number of times compounded per year, and t is the number of years.

6000 = 1000 (1+ 0.03/3)^3t 1) Divide both sides by 1000 to isolate the exponential part of equation

6 = (1 + 0.03/3)^3t 2) Simplify inside parenthesis

6 = (1.01)^3t 3) Take the log of both sides

Log₁₀6 = Log₁₀1.01^3t 4) Use exponent log property and move 3t to coefficient

Log₁₀6 = (3t) Log₁₀1.01 5) Divide both sides by 3Log₁₀1.01 to solve for t

t = Log₁₀6 / (3 Log₁₀1.01) 6. Use calculator

t = 60.02 years (60 years will be slightly less than $6000 total so you always need to round up to next compounding period which is 60 years and 4 months)

b) A = Pe^rt where A is the final amount of money, P is the initial investment, r is the rate (need to convert to decimal), and t is the number of years.

6000 = 1000e^0.03t 1) Divide both sides by 1000 to isolate the exponential part of equation

6 = e^0.03t 2) Take natural log of both sides

Ln(6) = Ln(e)^0.03t 3) Use exponent log property and move 0.03t to coefficient

Ln(6) = 0.03t Ln(e) 4) Ln(e) simplifies to 1 ( log_ee = 1 because e¹ = e, any logarithm with the same base and argument is 1)

Ln(6) = 0.03t 5) Divide both sides by 0.03

Ln(6) /0.03 = t 6) Use calculator

t = 59.73 years (This is continuous interest so there is no need to round up unless the problem specifies to the nearest year or month or something. Always round up like in previous problem)

Hope this helped, have a good day