Answer:
Given that:
The equation for the future value of a deposit earning compound interest is equation:
[tex]V(t) = P(1+\frac{r}{n})^{nt}[/tex] .....[1]
where,
P = the initial deposit
t = years invested
r = rate at which interest is compounded annually
.
n = number of times the interest is compounded per year
As per the statement:
After 10 years, a $2,000-dollar investment compounded annually has grown to $3600.
⇒P = $2000 and V(t) = $3600
Substitute in [1] we have;
[tex]3600 = 2000(1+\frac{r}{1})^{10 \cdot 1}[/tex]
Divide both sides by 2000 we have;
[tex]1.8 = (1+r)^{10}[/tex]
Taking log base 10 both sides we have;
[tex]\log_{10} 1.8 =\log_{10} (1+r)^{10}[/tex]
⇒[tex]0.255272505 = 10 \log_{10} 1+r[/tex]
Divide both sides by 10 we have;
[tex]0.0255272505 =\log_{10} 1+r[/tex]
⇒[tex]10^{0.0255272505} = 1+r[/tex]
Simplify":
[tex]1.06= 1+r[/tex]
Subtract 1 from both sides we have;
[tex]0.06=r[/tex]
or
r = 0.06 = 6%
Therefore, 6% is the interest rate to the nearest whole-number percent
