you want (1+r)^n = 2
so, for example, if n=5, you want (1+r)^5 = 2, so r = 14.87%
Determine an interest rate less than 15%, a period of investment greater than two years, and a regular payment that will result in the total amount of interest you earn being equal to the total amount of money you put in? (for example, under what conditions will you have a future value of $1 000 000, having earned $500 000 interest?)
2 answers
In effect what you are asking is, what is the relation between time and interest rate for money to double.
Before the ease of calculations, we used a procedure called the "rule of 72" which gave an approximation for that effect.
e.g. for a rate of 6% it would take appr 12 years for money to double, 6*12 = 72
check: 1.06^12 = 2.012 , almost doubled
or , 8% and 9 years, 1.08^9 = 1.999
so let's take your max of 15%, it would take 72/15 or 4.8 years
check : 1.15^4.8 = 1.96
and at the low end of 2 year, the rate would be 72/2 or 36% according to that 72 run, but that is way too high.
oobleck kicks in here with (1+r)^n = 2
pick any n>2 and you can find r
pick any r < .15 and you can find n
e.g. If r = .12 or 12%
1.12^n = 2
n = log2/log1.12 = 6.11 or about 6 years
Notice that still validates the rule of 72, 6*12 = 72
Before the ease of calculations, we used a procedure called the "rule of 72" which gave an approximation for that effect.
e.g. for a rate of 6% it would take appr 12 years for money to double, 6*12 = 72
check: 1.06^12 = 2.012 , almost doubled
or , 8% and 9 years, 1.08^9 = 1.999
so let's take your max of 15%, it would take 72/15 or 4.8 years
check : 1.15^4.8 = 1.96
and at the low end of 2 year, the rate would be 72/2 or 36% according to that 72 run, but that is way too high.
oobleck kicks in here with (1+r)^n = 2
pick any n>2 and you can find r
pick any r < .15 and you can find n
e.g. If r = .12 or 12%
1.12^n = 2
n = log2/log1.12 = 6.11 or about 6 years
Notice that still validates the rule of 72, 6*12 = 72
2 answers