possible

At the end of $t$ years, the future value of an investment of $\mathrm{\$}50,000$ at $6\mathrm{\%}$, compounded annually, is given by $S=50,000(1+0.06{)}^t$. In how many years will the investment grow to $\mathrm{\$}106,646.41$ ?

The investment will grow to $\mathrm{\$}106,646.41$ in approximately $◻$ 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 $\mathrm{\$}106646.41$ in approximately $\boxed{{\displaystyle 13}}$ years