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

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.
