# Time required to make money 6 times of itself

A sum doubles itself in one year at a certain rate of interest, compounded annually. In how many years will a sum become six times itself under the same investment scheme?

I got confused in this simple problem because of two approaches I can think of -

First - Clearly rate of interest = 100%, So $$6P = P(1+1)^n$$ solving We get n = log6/log2, n = 2.58 Years

Second - At the end of 2 years Amount = 4P, now to convert 4P to 6P we need say x months, then $$2P=(4P*1*x)/12$$; x = 6 months, hence answer = 2.5 years

Why the difference?

• After one year it is $2$ times the original amount. After two years it is $4$ times the original amount. After four years it is $8$ times the original amount. So it is never exactly $6$ times the original amount. But your first argument is better than your second and is what I would use if compounding took place at shorter periods with an annual effective rate of $100\%$ – Henry Nov 14 '14 at 20:13
• @Henry, but what if a bank works like this, and you decide to withdraw money at the end of 2.5 years, what amount will you get? – PRYM Nov 14 '14 at 20:15
• If it works like the question, you will get $4$ (you miss the bonus coming at the end of the third year. But if it works like a British bank and it was foolish enough to pay $100\%$ APR then you get something like $2^{2.5} \approx 5.65$ – Henry Nov 14 '14 at 20:25

The amount of money after $x$ year ($x$ a real number) is not multiplied by $(1+r)^x$. Because the compound is not continuous. The amount invested after $k\in \Bbb N$ years is multiplied by $(1+r)^x$ and this fixed amount is invested until the end of the year, and only then the interests are reinvested.
This leads to the multiplier (with $E(x)$ the integer part of $x$): $$(1+r)^{E(x)}\times (1 + r(x-E(x)))$$ (the last factor is for the year not yet finished).
In your case this gives $$2^{E(x)}\times (1 + {x-E(x)}) = 6$$
which gives $E(x) = 2$ and $$1 + {x-2} = \frac 64 = 1.5\implies x = 2.5$$