Question 1197961

How long would it take to double $2000 in the bank deposit with 4% interest compounded monthly
<pre>Since this refers to DOUBLING, the initial amount is never needed, as the formula is always: {{{matrix(1,3, 2, "=", (1 + i/m)^mt)}}}
This gives us: {{{matrix(1,3, 2, "=", (1 + .04/12)^(12t))}}} ------ Substituting .04 for i (interest), and 12 for m (annual compounding periods)
             {{{matrix(1,3, 12t, "=", log ((1 + .04/12), (2)))}}} ----- Converting to LOGARITHMIC form
Time it'll take to double, or {{{highlight_green(matrix(1,11, t, "=", (log ((1 + .04/12), (2)))/12, or, (log ((2))/log((1 + .04/12))) * (1/12), or, log ((2))/12log((12.04/12)), "=", "17.35754463,", or, matrix(1,5, 17, "years,", 4.29, months, "(approximately)")))}}} 
However, since this investment is being compounded MONTHLY, you need to ROUND UP to the next month, which is month 5. 
So, correct answer should be 17 years, 5 months.

A QUICKER method is to use the Rule of 69.3. This rule is MOST PRECISE for interest rates below 6%. From 6% - 10%, the MOST PRECISE
is the Rule of 72. Depending on the interest rate, other rules (Rule of 70 & 78) can be used to get the most accurate time-estimate. 

Using the Rule of 69.3, we get: i (in percent form) * time (in years) = 69.3
                                                                   4t = 69.3 ------- Substituting 4 for interest rate
                                          Time taken to double, or {{{matrix(1,6, t, "=", 69.3/4, "=", 17.3, years)}}}.

                                               See how precise the Rule of 69.3 is as a time-predictor?</pre>