Answer:
[tex]17.4\ 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=\$3,000\\ r=0.04\\n=12\\ A=\$6,000[/tex]
substitute in the formula above
[tex]\$6,000=\$3,000(1+\frac{0.04}{12})^{12t}[/tex]
[tex]2=(\frac{12.04}{12})^{12t}[/tex]
Applying log both sides
[tex]log(2)=log[(\frac{12.04}{12})^{12t}][/tex]
[tex]log(2)=(12t)log[(\frac{12.04}{12})][/tex]
[tex]t=log(2)/[(12)log(\frac{12.04}{12})]=17.4\ years[/tex]
