how much will $\$10,000$ grow to in three years at the same rate of interest?

Question

If $\$64$ grows to $\$128$ in four years at a constant effective annual interest rate, how much will $\$10,000$ grow to in three years at the same rate of interest?

I tried to use $(1+i/n)^{n-1}$ to solve for the interest rate but I don't understand how to use the numbers $64$ and $128$...

I know the answer is $\$16,817.93$

thanks in advance

Answer 1

Let's assume interest rate $i$, where $i=0.1$ means $i = 10\%$

Now, if you start with amount $x$, then after 1 year you get as interest $i*x$, which means that after one year you have $x + i*x = (1 + i) * x$

After the second year, you earn interest over that amount, so you get $i*(1+i)*x$ interest. Hence, after two years you have amount $(1+i) * x + i * (1 + i) * x = (1 + i) * (1 + i) * x = (1 + i) ^ 2 * x$

You can now hopefully see that this pattern repeats: after $n$ years, you will have amount $(1 + i)^n*x$

So, in your case, since you start with $x = 64$, and since after 4 years you are at 128, you need to solve for: $64*(1 + i)^4 = 128$, i.e. for: $(1 + i)^4=2$ to find interest rate $i$. So: $ 1 + i = 2^{\frac{1}{4}}$

So, $i = 2^{\frac{1}{4}} - 1 \approx 0.189$ or about $18.9%$ (hmm, sounds like a typical credit card ...)

Anyway, when you start with $x= 10,000$, after three years you will have $10,000*(1+i)^3$. So, here you can fill in $i$ ... or you can you the earlier finding that $ 1 + i = 2^{\frac{1}{4}}$, and hence $(1+i)^3 = 2^{\frac{3}{4}}$. Either way, you get $10,000*(1+i)^3 \approx 16,818$

Answer 2

Note that $128 = 2\cdot 64$. So, the amount of money doubles every $4$ years, giving us $$P(t) = P_0 2^{t/4}$$ Where $P(t)$ is the amount of money at time $t$ years. In this question, $P_0 =10000$, and so $$P(3) = 10000\cdot 2^{3/4} \approx 16818$$