Compound interest with a compounding interest rate

Question

I have an investment which pays 3% interest (r) annually but it also increases the interest rate every year by 5% (g). I re-invest all interest payments at the start of each year. How many years (t) will it take for the future value (FV) to be double my initial investment?

E.g. starting with an initial value of 1000:

Year 1 : Yield = 3.00%, FV = 1000 + (1000 * 3%) = 1030.00
Year 2 : Yield = 3.15% (3% increased by 5%), FV = 1030 + (1030 x 3.15%) = 1062.45
Year 3 : Yield = 3.31% (3.15% increased by 5%), FV= 1062.45 + (1062.45 x 3.31%) = 1097.59
Year 16 : FV = 2001.13

I can calculate the answer manually in a spreadsheet but I would like to know how to solve this mathematically and obtain an expression for t in terms of r & g?

Answer 1

Starting with an initial rate $r$ ($r=.03$ in the example) and increasing rate $g$ ($g=.05$ in the example) then the compounding gives the formula $P_{k+1}=P_k(1+r(1+g)^k)=P_k*N_k$ after $k$ periods. So you get a factor at $k$ periods $M_k=\prod_{t=0}^{k-1} N_t=\prod_{t=0}^{k-1} (1+r(1+g)^{t})$. Now you want to solve for $k$ so that $M_k=2$. Using logarithms $ln(1+x)\approx x$ you could approximate this to solve for $k$ in $\ln(2)=\sum_{t=0}^{k-1} r(1+g)^t=((1+g)^{k}-1)\frac{r}{g}$ using logarithms again.

For your initial parameters this gives $k=15.74$.

Answer 2

$$PI(1+iG)(1+iG^2)(1+iG^3)\cdots(1+iG^{(n-1)}) = 2P$$

Taking Log on both sides

$$lnI + ln(1+iG) + ln(1+iG^2)+\cdots+ln(1+iG^{(n-1)}) = ln2$$

Using the approximation $ln(1+x) \approx x$,

$$iG(1+G+G^2+\cdots+G^{(n-2)}) = ln\left(\frac{2}{I}\right)$$

$$\frac{(G^{(n-1)} - 1}{g} = \frac{ln(\frac{2}{I})}{iG}$$

$$G^{(n-1)} =1+\frac{gln(\frac{2}{I})}{iG}$$

Taking logarithm again

$$n = 1+\dfrac{ln(1+\frac{gln(\frac{2}{I})}{iG})}{lnG}$$

For your values of i = 0.03 and g = 0.05, the value is $15.74591277$ years.