In your question you have small inconsistency. First $A$ and $D$ are defined as two units of measurement for film sensitivity, so "21 DIN are equivalent to 100 ASA" could be rewritten as $21\ D=100\ A$ (like equation $1\ [inch] = 0.0254\ [meters]$). Then in equation $D=1+10\ \log_{10}(A)$ $A$ and $D$ are clearly real variables - contradiction! I will stick to former convention, then correct relation between ASA and DIN would be $$x\ A=(1+10\ \log_{10}(x))\ D$$ This is helpful to calculate DIN, when ASA value $x$ is given. Now doubling ASA yields: $$2x\ A=(1+10\ \log_{10}(2x))\ D$$ and we can apply logarithm identity $ \log_b(xy)=\log_b(x)+\log_b(y)$ to get $$1+10\ \log_{10}(2x)=1+10\ \log_{10}(x)+10\ \log_{10}(2)$$ $$2x\ A=(1+10\ \log_{10}(x)+10\ \log_{10}(2))\ D$$ so doubling ASA at any value $x$ increases DIN by constant $10\ \log_{10}(2) \approx 3$.
Sidenote: equation can be reversed to obtain ASA from DIN: $$ y\ D = 10^{\frac{y-1}{10}}\ A $$