Solution
The given equation to get the accumulated amount is
[tex]\begin{gathered} A=Pe^{rt} \\ \text{Where r = rate = 10\%}=\frac{10}{100}=\text{ 0}.1 \\ t\text{ = time in years} \\ P\text{= Amount invested}=\text{ \$6000} \\ A=\text{ Accumulated amount = 2 }\times6000\text{ = \$12000 } \end{gathered}[/tex]
Therefore, by substituting in these values, t will be given as
[tex]\begin{gathered} 12000\text{ = 6000}\times e^{0.1\times t} \\ \frac{12000}{6000}=e^{0.1t} \\ 2=e^{0.1t} \\ \ln \text{ 2 = 0.1t} \\ 0.6931471806=0.1t \\ t\text{ =}\frac{0.6931471806}{0.1} \\ t\text{ = }6.931471806 \\ t\approx6.9\text{ years to 1 decimal place} \end{gathered}[/tex]
t approximately = 6.9 years to 1 decimal place.
