How do you use natural log to solve for x?
i really have no clue!
$$(1/16)^{3x+1} = (1/18)^{x+1}$$ Take the natural log on both sides; we will get $$\ln 16^{-3x-1}=\ln 18^{-x-1}.$$ Using the properties for $\ln$, we have $$(-3x-1) \ln 16 = (-x-1) \ln 18.$$ It is easy to solve for $x$ now. Good luck!
Lets continue from measure2012's answer (you don't really need long division). Define $a = \mathrm{ln} 16$ and $b = \mathrm{ln} 18$. Then the following are equivalent.
If you want to eliminate $b$ and $a$ in this final answer, you can do that by replacing them with $\mathrm{ln} 16$ and $\mathrm{18}$. However, its probably a bit pointless.