Step 1 :Suppose that \(P\) is invested in a savings account in which interest, \(k\), is compounded continuously at 5% per year. The balance \(P(t)\) after time \(t\), in years, is given by the formula \(P(t)=P e^{k t}\).
Step 2 :The exponential growth function in terms of \(P\) and 0.05 is \(P(t)=P e^{0.05 t}\).
Step 3 :If $6000 is invested, we can find the balance after 2 years by substituting \(P=6000\), \(t=2\), and \(k=0.05\) into the equation.
Step 4 :Calculating \(P(t)=6000 e^{0.05 \times 2}\) gives a balance of approximately 6631.03.
Step 5 :\(\boxed{6631.03}\) is the balance after 2 years when $6000 is invested.