How long does it take for an investment to double in value if it is invested at 14\% compounded quarterly and compounded continuously? a) At 14\% compounded quarterly, the investment doubles in how many years? b) At 14\% compounded continuously, the investment doubles in how many years?

Step 1
Compound interest:
In compound interest, interest is added back to the principal sum so that interest is earned on that added during the next compounding period. That is, compound interest will give an interest on the interest. The interest payments will change in the time period in which the initial sum of money stays in the bank or with the barrower.
The general formula for compound interest is,
$A=P\cdot (1+r/n{)}^{nt}$
Where:
A is the future value of the investment loan including the loan,
P is the principle amount,
r is the annual interest rate in decimals,
n is the number of times interest is compounded per year,
t is the time of years the money is invested or borrowed.
Step 2
a) Find the number of years in which the investment will be doubled at $14\mathrm{\%}$ interest compounded quarterly:
The aim is to double the invested or principal amount at the given interest rate.
The future value of the investment loan including the loan should be the double of principal amount at $14\mathrm{\%}$ interest compounded quarterly.
Here,
Let the principal amount or the invested amount is P.
The future value of the invested amount including the amount is $A=2P$
Annual interest rate is $r=14\mathrm{\%}=0.14$
The number of times interest is compounded per year is quarterly. That is, $n=4.$
The number of years required to double the invested money is invested t.
The number of years in which the investment will be doubled at $14\mathrm{\%}$ interest compounded quarterly is obtained as 5.04 years from the calculation given below:
$A=P\times {(1+\frac{r}{n})}^{nt}$
$2P=P\times {(1+\frac{0.14}{4})}^{4t}$
$2=(1+0.035{)}^{4t}$
$2=(1.035{)}^{4t}$
Take natural logaritm on both sides
$\ln (2)=\ln ({1.035}^{4t})$
$=4t\ln (1.035)$
$t=\frac{\ln (2)}{4\times \ln (1.035)}$
$=5.04$
Step 3 Continuous compound interest:
In compound interest, interest is added back to the principal sum so that interest is earned on that added during the next compounding period. That is, compound interest will give an interest on the interest.
In continuous compound interest, the principal amount will be constantly earning interest and the interest keeps earning on the interest earned.
The general formula for continuous compound interest is,
$A=P\cdot e^{rt}$
Where:
A is the future value of the investment loan including the loan,
P is the principle amount,
r is the interest rate in decimals,
t is the time of years the money is invested or borrowed.
Step 4
b) Find the number of years in which the investment will be doubled at $14\mathrm{\%}$ interest compounded continuously:
The aim is to double the invested or principal amount at the given interest rate.
The future value of the investment loan including the loan should be the double of principal amount at $14\mathrm{\%}$ interest compounded continuously.
Here,
Let the principal amount or the invested amount is P.
The future value of the invested amount including the amount is $A=2P$
Annual interest rate is $r=14\mathrm{\%}=0.14,$
The number of years required to double the invested money is invested t.
The number of years in which the investment will be doubled at $14\mathrm{\%}$ interest compounded continuously is obtained as 4.95 years from the calculation given below:
$A=P\times e^{rt}$
$2P=P\times e^{rt}$
$2=e^{rt}$
Take natural logarithm on both sides
$\ln (2)=\ln (e^{rt})$
$\ln (2)=rt$
$t=\frac{\ln (2)}{r}$
$=\frac{\ln (2)}{0.14}.[\because r=14\mathrm{\%}=0.14]$
$=4.95$
Step 5
Answer: a) In 5.04 years, the investment will be doubled at $14\mathrm{\%}$ interest compounded quarterly.
b) In 4.95 years, the investment will be doubled at $14\mathrm{\%}$ interest compounded continuously.