Answer:
[tex]t=10.2\ years[/tex]
Step-by-step explanation:
we know that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
[tex]t=?\ years\\P=\$2,000\\A=\$3,000\\ r=0.04\\n=12[/tex]
substitute in the formula above
[tex]3,000=2,000(1+\frac{0.04}{12})^{12t}[/tex]
[tex]1.5=(\frac{12.04}{12}})^{12t}[/tex]
Applying log both sides
[tex]log(1.5)=log[(\frac{12.04}{12}})^{12t}][/tex]
[tex]log(1.5)=(12t)log(\frac{12.04}{12})[/tex]
[tex]t=log(1.5)/[(12)log(\frac{12.04}{12})][/tex]
[tex]t=10.2\ years[/tex]
