Respuesta :
Answer:
For an interest of 13%, the amount of time it will take the principal to double is:
Annually: About 5.67 years.
Monthly: About 5.36 years.
Daily: About 5.33 years.
Continuously: About 5.33 years.
Step-by-step explanation:
The standard formula for compound interest is given by:
[tex]\displaystyle A=P(1+\frac{r}{n})^{nt}[/tex]
Where P is the principal, A is the amount afterwards, r is the interest rate, n is the amount of times it is compounded per year, and t is the number of years.
We want to find t for n=1 (annually), n=12 (monthly), n=365 (daily) and continuously when our final amount is double our principal.
In other words, A=2P. Therefore:
[tex]\displaystyle 2P=P(1+\frac{r}{n})^{nt}[/tex]
We can divide both sides by P:
[tex]\displaystyle 2=(1+\frac{r}{n})^{nt}[/tex]
Since our rate is 13%, r=0.13. Hence:
[tex]\displaystyle 2=(1+\frac{0.13}{n})^{nt}[/tex]
We now have our general formula which we can use to solve.
Annually:
For an annual compound, n=1. Therefore:
[tex]\displaystyle 2=(1+\frac{0.13}{(1)})^{(1)t}[/tex]
Simplify :
[tex]2=(1.13)^t[/tex]
We can take the log (base 10) of both sides:
[tex]\log2=\log1.13^t[/tex]
We can move the t in front:
[tex]\log2=t\log1.13[/tex]
Therefore:
[tex]\displaystyle t=\frac{\log 2}{\log 1.13}\approx5.67[/tex]
So, for an annual compound, it will take about 5.67 years for the principal to double.
Monthly:
For monthly compounds, n=12. Therefore:
[tex]\displaystyle 2=(1+\frac{0.13}{(12)})^{(12)t}[/tex]
We can take the log of both sides:
[tex]\displaystyle \log 2=\log{(1+\frac{0.13}{12})^{12t}}[/tex]
Again, we can move the 12t to the front:
[tex]\displaystyle \log 2=12t\log{(1+\frac{0.13}{12})[/tex]
Therefore:
[tex]\displaystyle t=\frac{\log2}{12\log{(1+\frac{0.13}{12})}}\approx5.36[/tex]
So, for monthly compounds, it will take about 5.36 years for the principal to double.
Daily:
For daily compounds, n=365. Therefore:
[tex]\displaystyle 2=(1+\frac{0.13}{365})^{365t}\\[/tex]
Take the log of both sides and moving the 365t to the front yields:
[tex]\displaystyle \log{2}=365t \log(1+\frac{0.13}{365})[/tex]
Hence:
[tex]\displaystyle t = \frac{ \log 2}{365\log(1+\frac{0.13}{365} ) } \approx5.33[/tex]
So, for daily compounds, it will take about 5.33 years for the principal to double.
Continuously:
For continuous compound, we will need to use the continuous compound formula:
[tex]A=Pe^{rt}[/tex]
So, A=2P:
[tex]2P=Pe^{rt}[/tex]
Dividing both sides by P:
[tex]2=e^{rt}[/tex]
Since r=13% or 0.13:
[tex]2=e^{0.13t}[/tex]
This time, we can take the natural log of both sides:
[tex]\ln2=\ln e^{0.13t}[/tex]
So:
[tex]\ln 2 = 0.13t \ln e[/tex]
The natural log of e is simply 1. Therefore:
[tex]\ln 2=0.13 t[/tex]
Hence:
[tex]\displaystyle t=\frac{\ln 2}{0.13}\approx5.33[/tex]
For, for continuously compounded, it will take about 5.33 years for the principal to double.
