jmc

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 $x\log f(r)=\log \frac{7}{3}$. 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 $\boxed{\text{(B)}}3(1+r/10{)}^x=7$.