If $\log_{10}2=a$ and $\log_{10}3=b$, then $\log_{5}12=$?
$\textbf{(A)}\ \frac{a+b}{a+1}\qquad \textbf{(B)}\ \frac{2a+b}{a+1}\qquad \textbf{(C)}\ \frac{a+2b}{1+a}\qquad \textbf{(D)}\ \frac{2a+b}{1-a}\qquad \textbf{(E)}\ \frac{a+2b}{1-a}$
From the [change of base formula](https://artofproblemsolving.com/wiki/index.php/Change_of_base_formula), $\log_{5}12 = \frac{\log_{10}12}{\log_{10}5}$.
For the numerator, $2 \cdot 2 \cdot 3 = 12$, so $\log_{10}12 = \log_{10}2 + \log_{10}2 + \log_{10}3 = 2a+b$.
For the denominator, note that $\log_{10}10 = 1$.  Thus, $\log_{10}5 = \log_{10}10 - \log_{10}2 = 1 - a$.
Thus, the fraction is $\frac{2a+b}{1-a}$, so the answer is $\boxed{\textbf{(D)}}$.