We can use the future value formula to solve the exercise.
[tex]\begin{gathered} FV=PV(1+\frac{r}{n})^{nt} \\ \text{ Where} \\ FV\text{ is the future value} \\ PV\text{ is the present value} \\ \text{r is the yearly rate} \\ t\text{ is the number of years the money is invested} \end{gathered}[/tex]In this case, we have:
[tex]\begin{gathered} FV=6600 \\ PV=3000 \\ r=7.5\%=\frac{7.5}{100}=0.075 \\ n=2 \\ t=? \end{gathered}[/tex][tex]\begin{gathered} FV=PV(1+\frac{r}{n})^{nt} \\ 6,600=3,000(1+\frac{0.075}{2})^{2t} \\ 6,600=3,000(1+0.0375)^{2t} \\ 6,600=3,000(1.0375)^{2t} \\ \text{ Divide by 3000 from both sides} \\ \frac{6,600}{3,000}=\frac{3,000(1.037,5)^{2t}}{3,000} \\ 2.2=(1.0375)^{2t} \\ \text{ Apply natural logarithm from both sides} \\ \ln(2.2)=\ln((1.0375)^{2t}) \\ \text{ Apply the power property of natural logarithms }\ln(m^p)=p\ln(m) \\ \operatorname{\ln}(2.2)=2t\operatorname{\ln}(1.0375) \\ \text{ Divide by }2\operatorname{\ln}(1.0375)\text{ from both sides} \\ \frac{\operatorname{\ln}(2.2)}{2\operatorname{\ln}(1.0375)}=\frac{2t\operatorname{\ln}(1.0375)}{2\operatorname{\ln}(1.0375)} \\ 10.7\approx t\Rightarrow\text{ The symbol }\approx\text{ is read 'approximately'} \end{gathered}[/tex]AnswerThe number of years is 10.7.
