If the bases in a quotient or product are equivalent, you can manipulate exponents (subtracting for in the case of quotients, adding in the case of products). If two distinct bases have equivalent exponents, you can manipulate the bases: for $\dfrac{a^7}{b^7} = \left(\dfrac ab\right)^7$ and $a^7\cdot b^7 = (a\cdot b)^7$
For your problem:
$\frac{2^{128}}{2\cdot10^{19}} = \frac{2^{(128-1)}}{10^{19}}= \frac{2^{127}}{10^{19}} = \frac{2^{127}}{(2 \cdot 5)^{19}}= \frac{2^{127}}{2^{19}\cdot 5^{19}} = \frac{2^{(127 - 19)}}{5^{19}} = \frac{2^{108}}{5^{19}}$
That's as simpified as it gets, without computing $2^{108}$ and $5^{19}$, since $2$ and $5$ are both prime (hence coprime!).
If you meant to write: $\dfrac{2^{128}}{(2\cdot10)^{19}}$, then: $\frac{2^{128}}{(2\cdot10)^{19}} = \frac{2^{128}}{2^{19}\cdot 10^{19}} = \frac{2^{128}}{2^{19}\cdot 2^{19} \cdot 5^{19}} = \frac{2^{128- 19 - 19}}{5^{19}} = \frac{2^{90}}{5^{19}}$
As you see, there is a very big difference between: $\quad \dfrac{2^{128}}{(2\cdot10)^{19}}\quad $ and $\quad\dfrac{2^{128}}{2\cdot10^{19}}$