Kevin deposited $\$ 3000$ into an account with a $3.4 \%$ annual interest rate, compounded quarterly. Assuming that no withdrawals are made, how long will it take for the investment to grow to $\$ 3237$ ?
Do not round any intermediate computations, and round your answer to the nearest hundredth.

Step 1 ：Given that Kevin deposited $3000 into an account with a 3.4% annual interest rate, compounded quarterly. We are to find how long it will take for the investment to grow to $3237.

Step 2 ：We can use the formula for compound interest, which is \(A = P(1 + \frac{r}{n})^{nt}\), where:

Step 3 ：\(A\) is the amount of money accumulated after n years, including interest.

Step 4 ：\(P\) is the principal amount (the initial amount of money).

Step 5 ：\(r\) is the annual interest rate (in decimal).

Step 6 ：\(n\) is the number of times that interest is compounded per year.

Step 7 ：\(t\) is the time the money is invested for in years.

Step 8 ：In this case, we know the following:

Step 9 ：\(A = \$3237\)

Step 10 ：\(P = \$3000\)

Step 11 ：\(r = 3.4\% = 0.034\)

Step 12 ：\(n = 4\) (since interest is compounded quarterly)

Step 13 ：We need to solve for \(t\). We can rearrange the formula to solve for \(t\) as follows:

Step 14 ：\(t = \frac{\log(\frac{A}{P})}{n \cdot \log(1 + \frac{r}{n})}\)

Step 15 ：Substituting the known values into the equation, we get:

Step 16 ：\(t = \frac{\log(\frac{3237}{3000})}{4 \cdot \log(1 + \frac{0.034}{4})}\)

Step 17 ：Solving the equation, we find that \(t = 2.2458052304618548\)

Step 18 ：Rounding to the nearest hundredth, we get \(t = 2.25\)

Step 19 ：Final Answer: It will take approximately \(\boxed{2.25}\) years for the investment to grow to $3237.