Answer:
After 7.1 years
Step-by-step explanation:
The compound interest formula is given by:
[tex]A(t) = P(1 + \frac{r}{n})^{nt}[/tex]
Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit year and t is the time in years for which the money is invested or borrowed.
In this question:
We have to find t, for which [tex]A(t) = 3000[/tex] when [tex]P = 2000, r = 0.0575, n = 4[/tex]
So
[tex]A(t) = P(1 + \frac{r}{n})^{nt}[/tex]
[tex]3000 = 2000(1 + \frac{0.0575}{4})^{4t}[/tex]
[tex](1.014375)^{4t} = \frac{3000}{2000}[/tex]
[tex](1.014375)^{4t} = 1.5[/tex]
[tex]\log{(1.014375)^{4t}} = \log{1.5}[/tex]
[tex]4t\log{1.014375} = \log{1.5}[/tex]
[tex]t = \frac{\log{1.5}}{4\log{1.014375}}[/tex]
[tex]t = 7.1[/tex]
After 7.1 years
