Solution:
From the amortization function:
[tex]\begin{gathered} A(t)=A_o(1+\frac{r}{n})^{nt} \\ where \\ A_o\Rightarrow initial\text{ amount invested} \\ r\Rightarrow annual\text{ interest rate} \\ n\Rightarrow number\text{ of times compounded} \\ t\Rightarrow time \end{gathered}[/tex]
Given that $10,000 is invested at an annual rate of 5% compounded quarterly, this implies that
[tex]\begin{gathered} A_o=10000 \\ r=5\%=0.05 \\ n=4 \end{gathered}[/tex]
Thus, by substituting these values into the above equation, we have
[tex]\begin{gathered} A(t)=10000(1+\frac{0.05}{4})^{4t} \\ \Rightarrow A(t)=10000(1.0125)^{4t} \end{gathered}[/tex]
Given that the amount earned is double the initial amount invested, this implies that
[tex]A(t)=20000[/tex]
Thus, we have
[tex]20000=10000(1.0125)^{4t}[/tex]
To solve for t,
[tex]\begin{gathered} divide\text{ both sides by 10000} \\ \frac{20000}{10000}=\frac{10000(1.0125)^{4t}}{10000} \\ \Rightarrow2=(1.0125)^{4t} \\ take\text{ the logarithm of both sides,} \\ \ln2=\ln(1.0125)^{4t} \\ 0.6931471806=4t\times0.01242252 \\ \Rightarrow4t=55.79763048 \\ divide\text{ both sides by the coefficient of t, which is 4} \\ thus, \\ \frac{4t}{4}=\frac{55.79763048}{4} \\ \Rightarrow t=13.94940762 \end{gathered}[/tex]
Hence, it takes 13.94940762 years to double the amount invested.
