smc

(B) Intuitively, $x$ will be largest for that 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. And 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 $r^2<10$, dividing both sides by $10r$ we get $r/10<1/r$. And finally, $r/100<1$, therefore $r^2/100<r$, and as both sides are positive, we can take the square root and get $r/10<\sqrt{r}$. Thus the answer is $\boxed{3(1+r/10{)}^x=7}$.