Respuesta :
Answer:
[tex]t \approx 10.15351... \ (years)[/tex]
Step-by-step explanation:
Interest compound is defined as
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
Where,
[tex]P[/tex] is the principal.
[tex]r[/tex] is the rate in decimal.
[tex]n[/tex] is the number of compounded periods within a year.
[tex]t[/tex] is the time in years.
In this case, we have
[tex]P=2000\\A=3000\\r=4\%=0.04\\ n=12\\t=?[/tex]
Replacing all values, we have
[tex]3000=2000(1+\frac{0.04}{12})^{12t}[/tex]
Now, we solve for [tex]t[/tex]
[tex]\frac{3000}{2000}=(1+ 0.003)^{12t}[/tex]
[tex]1.5=(1.003)^{12t}[/tex]
[tex]ln(1.5)=ln((1.003)^{12t} )\\ln(1.5)=12t \times ln(1.003)\\12t=\frac{ln(1.5)}{ln(1.003)} \\12t=\frac{0.4}{0.003}\\ 12t=133.33\\t=\frac{133.33}{12}\\ t \approx 11.11[/tex]
Our result is different because at each step we approximated results.
If you use a calculater, you would find a more exact result would be
[tex]t \approx 10.15351...[/tex]
Therefore, the right answer is the third choice.