So the way I tried to approach it was by making it $4^x = 7x -2$, and then I would know that there is an exponential curve and a straight line intersection.
The second derivative of the exponential curve would be always positive thus concave up, and then the straight line can intersect the boundary region of the concave up curve (convex shape) twice at most.
However, my professor asked me to use Calculus (Rolle's theorem and probably the Intermediate Value Theorem) in this case, and I am somehow lost in trying to figure that out.
Any help will be great. Thanks!
See, we have that $f(0) > 0$, $f(1) < 0$ and again $f(2) > 0$. Then by intermediate value theorem, you have two roots, one in $(0,1)$ other in $(1,2)$
Now look at the derivative function $f'(x) = 4^x \ln 4-7$. Thus you can show $f'(x) < 0$ for $x<0$. And $f'(x) > 0$ for $x > 2$.
Edit:
As John Lou says, we can show that $f'(x) = 4^x\ln 4 - 7 = 0$ has only one root as $4^x$ is increasing. Thus $f(x)$ cannot have more roots, because if we had more than two roots of $f(x)$, then $f'(x)$ would have had more roots (rolle's theorem).