Question Watch Video Show Examples A person invested $\$ 5,500$ in an account growing at a rate allowing the money to double every 14 years. How long, to the nearest tenth of a year would it take for the value of the account to reach $\$ 6,800$ ? Answer Attempt 1 out of 2
Solution
Step 1 ：The problem is about exponential growth, specifically about the time it takes for an investment to reach a certain value given a doubling time. The formula for exponential growth is \(A = P * 2^{(t/T)}\), where A is the final amount, P is the principal amount, t is the time, and T is the doubling time.
Step 2 ：We can rearrange this formula to solve for t: \(t = T * \log_2(A/P)\).
Step 3 ：Given that the principal amount (P) is \$5500, the final amount (A) is \$6800, and the doubling time (T) is 14 years, we can plug in these values into the formula.
Step 4 ：By substituting the given values into the formula, we get \(t = 14 * \log_2(6800/5500)\).
Step 5 ：Solving this equation gives us \(t \approx 4.285443788159517\).
Step 6 ：Rounding to the nearest tenth of a year, we get \(t \approx 4.3\) years.
Step 7 ：Final Answer: It would take approximately \(\boxed{4.3}\) years for the value of the account to reach \$6800.
