How long will it take for a $\$ 4000$ investment to grow to $\$ 7160$ at an annual rate of $10.8 \%$, compounded monthly? Assume that no withdrawals are made. Do not round any intermediate computations, and round your answer to the nearest hundredth.

Step 1 ：We are given the following values: Principal amount (P) = $4000, Final amount (A) = $7160, Annual interest rate (r) = 10.8% or 0.108 in decimal, and the number of times the interest is compounded per year (n) = 12.

Step 2 ：We need to find the time (t) it will take for the investment to grow from $4000 to $7160. We can use the formula for compound interest, rearranged to solve for t: \(t = \frac{\log(A/P)}{n \cdot \log(1 + r/n)}\)

Step 3 ：Substituting the given values into the formula, we get: \(t = \frac{\log(7160/4000)}{12 \cdot \log(1 + 0.108/12)}\)

Step 4 ：Solving the above expression, we find that \(t \approx 5.415108127511201\)

Step 5 ：Rounding to the nearest hundredth, we get \(t \approx 5.42\)

Step 6 ：So, the time it will take for a $4000 investment to grow to $7160 at an annual rate of 10.8%, compounded monthly, is approximately \(\boxed{5.42}\) years.