$f(x) = ar^x $ given that $r > 0$.
I'm given two points ($3$, $\frac{8}{9}$) and ($4$, $\frac{16}{27}$).
My textbook then says $$r = \frac{\frac{16}{27}}{\frac{8}{9}}$$
Why does this work? I can't figure out why $y_2/y_1 = r$.
There's some other similar questions asked on this site but the answers are too complicated, using integrals and natural logarithms and such to explain something this simple.
Like Mick said, all you're doing is solving the system $$ar^3=\frac89$$ $$ar^4=\frac {16}{27}$$
If you divide the second equation by the first, you get $$\frac{ar^4}{ar^3}=\frac{\frac{16}{27}}{\frac{8}{9}}$$ $$\Rightarrow r=\frac{\frac{16}{27}}{\frac{8}{9}}$$