The values of $k$ for which $ \log(2x) \leq kx \leq e^{x/2}$ for all $x > 0 $ So I'm trying to solve a system of equations and I checked some other guys solution and he divides the function by the derivate, like so: $f(x)/f'(x)$.
Find the values of the real constant $k$ for which this holds true:
$$
\log(2x) \leq kx \leq e^{x/2}, \quad x > 0
$$
I convert this into 2 systems of equations:
$$
f(x)=\log(2x)=kx \ and \ f'(x)=1/x=k  \\
f(x)=e^{x/2}=kx  \ and \ f'(x)=\frac{e^{x/2}}{2}=k
$$
Then I got stuck,checked some other guys solution and he did this: $f(x)/f'(x)$ for both equations. Why?
Sorry for the long winded explanation but I really can't figure out why it's $f(x)/f'(x)$.
Progress
Figured it out. Dividing $f(x)$ with $f'(x)$ is just another way of creating an expression for x and inserting it into the other equation in the system.
 A: You are given two functions, $f(x) = e^{x/2}$ and $g(x) = \log(2x)$ for $x > 0$. The problem is to find out which lines through the origin $h(x) = kx$ pass in between these curves for all values of $x > 0$. The first thing to do, is to sketch a plot of $f$ and $g$. This plot shows that there is indeed a gap through which straight lines through the origin will pass. 
The line with the maximum allowed value of k will pass just underneath $f$. Therefore we must find the point $(x_0, f(x_0))$ on $f$ where the tangent passes through the origin. Now the derivative of $f(x)$ is given by $f'(x) = (1/2)e^{x/2}$. Therefore the tangent for $x=x_0$ is given by the equation:
tangent to $f = (x_0, e^{x_0/2}) + p (1, (1/2)e^{x_0/2}) $ 
The intersection with the $y$-axis is obtained by setting $p = -x_0$. The corresponding $y$-coordinate becomes $(1 - x_0/2) e^{x_0/2}$. Therefore $x_0 = 2$ is the solution. Substituting this value in the tangent yields $k = e/2$. This is the upper bound for $k$. 
For $g(x)$ we find the derivative $g'(x) = 2/x.$ The tangent through a point $x0$ on $g$:
tangent to $g = (x_0, \log(2x_0)) + q  (1, 2/x_0)$
The intersection with the y-axis occurs when $q = -x_0$. The corresponding $y$-value is $\log(2x_0) - 2$. Solving yields $x_0 = (1/2)e^2$. Substitution in the tangent yields $k = 4/e^2$. This the lower bound for $k$.  
In summary, we find that the set of straight lines through the origin that pass in between $f$ and $g$ are given by $y = kx$ with $k$ in the interval $(4/e^2, e/2)$.   
