Intuitively, $ x $ will be largest for the option for which the value in the parentheses is smallest. Formally, first note that each of the values in parentheses is larger than $ 1 $. Now, each of the options is of the form $ 3f(r)^x = 7 $. This can be rewritten as $ x\log f(r) = \log\frac 73 $. As $ f(r)>1 $, we have $ \log f(r)>0 $. Thus $ x $ is the largest for the option for which $ \log f(r) $ is smallest. As $ \log f(r) $ is an increasing function, this is the option for which $ f(r) $ is smallest. We now get the following easier problem: Given that $ 0<r<3 $, find the smallest value in the set $ \{ 1+r, 1+r/10, 1+2r, 1+\sqrt r, 1+1/r\} $. Clearly, $ 1+r/10 $ is smaller than the first and the third option. We have $ r^2 < 10 $, so dividing both sides by $ 10r $, we get $ r/10 < 1/r $. Finally, $ r/100 < 1 $, therefore $ r^2/100 < r $. Since both sides are positive, we can take the square root and get $ r/10 < \sqrt r $. Thus the answer is $ \boxed{\text{(B)}} 3(1 + r/10)^x = 7 $.