Find an expression for $n$ in terms of $x$, $y$, and $z$.

Question

At a certain rate of compound interest, $100$ will increase to $200$ in $x$ years, $200$ will increase to $300$ in $y$ years, and $300$ will increase to $1,500$ in $z$ years. If $600$ will increase to $1,000$ in $n$ year, find an expression for $n$ in terms of $x$, $y$, and $z$.

I know:

$600(1+i)^n = 1000$

I wrote:

$200 = 100(1+i)^x$

$300 = 100(1+i)^{x+y}$

$1500 = 100(1+i)^{x+y+z}$

Also, I know that :

$600=100(1+i)^{2x+y}$

Hence:

$1000=100((1+i)^{x+y+z}-(1+i)^{x+z}-(1+i)^x)$

$1000=100(1+i)^{2x+y}((1+i)^{-x+z}-(1+i)^{-x}-(1+i)^{-x-y})$

$1000=600((1+i)^{-x+z}-(1+i)^{-x}-(1+i)^{-x-y})$

$(1+i)^n=(1+i)^{-x+z}-(1+i)^{-x}-(1+i)^{-x-y}$

I need to get $n$ as a function of $x$,$y$ and $z$

I have a problem with my last line. Does anyone knows a faster way to get the solution ?

Answer 1

Let's make it clean and simple. Instead of $1+i$, I'll put $q$ as the rate. And I'll cancel out all the unnecessary terms. So we have $$ 2 = q^x \qquad \frac{3}{2} = q^y \qquad 5 = q^z $$ Now we want to make the ratio $\frac{10}{6}$ with these. We can 'massage' the expression a bit: $$ \frac{10}{6} = \frac{2 \cdot 5}{ \left( \frac{3}{2} \right)\cdot 2 \cdot 2} = \frac{5}{ \left( \frac{3}{2} \right)\cdot 2} = 5 \cdot \left( \frac{3}{2} \right)^{-1} \cdot (2)^{-1} $$ Now you can just plug in the known values and you're done.