possible At the end of $t$ years, the future value of an investment of $\$ 50,000$ at $6 \%$, compounded annually, is given by $S=50,000(1+0.06)^{t}$. In how many years will the investment grow to $\$ 106,646.41$ ? The investment will grow to $\$ 106,646.41$ in approximately $\square$ years. (Do not round until the final answer. Then round to the nearest whole number as needed.)

Step 1 ：Given the future value of the investment \( S = 106646.41 \)

Step 2 ：Initial investment \( P = 50000 \)

Step 3 ：Annual interest rate \( r = 0.06 \)

Step 4 ：We need to find the number of years \( t \) such that \( S = P(1+r)^t \)

Step 5 ：Taking the natural logarithm of both sides of the equation to solve for \( t \)

Step 6 ：\( t = \frac{\ln(S) - \ln(P)}{\ln(1+r)} \)

Step 7 ：Substituting the given values \( t = \frac{\ln(106646.41) - \ln(50000)}{\ln(1+0.06)} \)

Step 8 ：Calculating the value of \( t \) gives \( t = 12.99999951606019 \)

Step 9 ：Rounding \( t \) to the nearest whole number gives \( t = 13 \)

Step 10 ：The investment will grow to \( \$ 106646.41 \) in approximately \(\boxed{13}\) years