jmc

We can write the given equation as $${\log }_{10}{2}^a+{\log }_{10}{3}^b+{\log }_{10}{5}^c+{\log }_{10}{7}^d=2005.$$Then $${\log }_{10}(2^a\cdot 3^b\cdot 5^c\cdot 7^d)=2005,$$so $2^a\cdot 3^b\cdot 5^c\cdot 7^d=10^{2005}.$

Since $a,$ $b,$ $c,$ $d$ are all rational, there exists some positive integer $M$ so that $aM,$ $bM,$ $cM,$ $dM$ are all integers. Then $$2^{aM}\cdot 3^{bM}\cdot 5^{cM}\cdot 7^{dM}=10^{2005M}=2^{2005M}\cdot 5^{2005M}.$$From unique factorization, we must have $aM=2005M,$ $bM=0,$ $cM=2005M,$ and $dM=0.$ Then $a=2005,$ $b=0,$ $c=2005,$ and $d=0.$ Thus, there is only $\boxed{1}$ quadruple, namely $(a,b,c,d)=(2005,0,2005,0).$