Assume $0 < r < 3$.  Below are five equations for $x$.  Which equation has the largest solution $x$?

$\textbf{(A)}\ 3(1 + r)^x = 7\qquad \textbf{(B)}\ 3(1 + r/10)^x = 7\qquad \textbf{(C)}\ 3(1 + 2r)^x = 7$ $\textbf{(D)}\ 3(1 + \sqrt {r})^x = 7\qquad \textbf{(E)}\ 3(1 + 1/r)^x = 7$
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 $\boxed{\text{(B)}} 3(1 + r/10)^x = 7$.