jmc

We first note that the $4^x$ term grows much faster than the other three terms. Indeed, since $x$ is a positive integer, we have: $$4^x<1^{x+2}+2^{x+1}+3^{x-1}+4^x<4^x+4^x+4^x+4^x=4^{x+1}.$$Thus we know that $1170$ lies between $4^x$ and $4^{x+1}$.

The first six powers of $4$ are $4^1=4,$ $4^2=16,$ $4^3=64,$ $4^4=256,$ $4^5=1024,$ and $4^6=4096.$

We can see that $1170$ lies between the last two of these, so we check $x=5$ and find: $$1^{x+2}+2^{x+1}+3^{x-1}+4^x=1+64+81+1024=1170.$$Therefore, $x=\boxed{5}$.