Value range of parameter inequality. Assume $a,b,c>0$ satisfies :$$5c-3a \leq b \leq 4c-a$$ and $$c\ln b \geq a + c\ln c$$. Find the value range of $\frac{b}{a}$.

My approach:
because $a>0$ and I want to construct the target expression : $\frac{b}{a}$. I divide the first inequality by $a$ in the three part. and I get : $$5\frac{c}{a}-3 \leq \frac{b}{a} \leq 4 \frac{c}{a} - 1$$. In the second inequality. I also want to construct the $\frac{b}{a}$. $$c\ln b -c \ln a \geq a+c\ln c -c \ln a \Rightarrow \frac{b}{a} \geq \frac{c}{a} \exp{\frac{a}{c}}$$
But I can't go any further. It is depressed me a lot ..  What is the right way to solve this ? I don't know..
 A: Here what I did:
We have the following inequality:
$$5c-3a\leq b\leq 4c-a.\quad\quad (E_0)$$
Since $b>0$, then I can divide by $b$:
$$\dfrac{5c-3a}{b}\leq 1\leq \dfrac{4c-a}{b}.\quad (1)$$
Further $(1)$ can be seen as:
$$\dfrac{5c-3a}{b}\leq 1\leq \dfrac{5c-3a+2a-c}{b},$$
or
$$\dfrac{5c-3a}{b}\leq 1\leq \dfrac{5c-3a}{b}+\dfrac{2a-c}{b},$$
or
$$0\leq 1-\dfrac{5c-3a}{b}\leq \dfrac{2a-c}{b}.\quad (2)$$
With $(2)$ we can conclude that: $\dfrac{2a-c}{b}\geq0$ and since $b>0$ then $2a-c\geq0$. 
Finally we have: $$\dfrac{c}{a}\leq2.\quad\quad (E_1)$$
Now, back to $(E_0)$ and divide it by $a$, we get:
$$5\dfrac{c}{a}-3\leq \dfrac{b}{a}\leq 4\dfrac{c}{a}-1.\quad (3)$$
Using $(E_1)$, we have an upper bound:
$$\dfrac{b}{a}\leq 4\dfrac{c}{a}-1\leq7.$$
Now, it time to use the logarithm inequality to get the lower bound.
We have:
$$c\log b\geq a+c\log c.$$
First divide it by $c$, we get:
$$\log b\geq \dfrac{a}{c}+\log c.$$
Again we use $(E_1)$ to see that: $\log \dfrac{b}{c}\geq \dfrac{a}{c} \geq\dfrac{1}{2}>0.$ (I needed this to guarantee that $\log \dfrac{b}{c}>0$). Then, this is useful to divide by $\log \dfrac{b}{c}$ in both sides. Then, 
$$\dfrac{c}{a}\geq\dfrac{1}{\log \dfrac{b}{c}}.$$ 
Back to $(3)$ and use the left hand side, we get:
$$-3+5\dfrac{c}{a}\geq -3+\dfrac{5}{\log \dfrac{b}{c}}.\quad (4)$$
Now divide $(E_0)$ by $c$ and apply log (we about the right hand side) you get:
$$\log \dfrac{b}{c}\leq \log(4-\dfrac{a}{c}).$$
Which is equivalent to (using $(E_1)$)
$$\log \dfrac{b}{c}\leq \log(4-\dfrac{a}{c})\leq \log\dfrac{7}{2}.$$
Now use $(4)$ to get:
$$-3+5\dfrac{c}{a}\geq -3+\dfrac{5}{\log \dfrac{b}{c}}\geq -3+\dfrac{5}{\log \dfrac{7}{2}}.\quad (E_2)$$
Conclusion:
$$-3+\dfrac{5}{\log \dfrac{7}{2}}\leq \dfrac{b}{a}\leq 7.$$
$$0.99\leq \dfrac{b}{a}\leq 7.$$
Edit: $\log$ is the natural logarithm. I calculated wrong before.
A: Let $B = \frac{b}a, \: C = \frac{c}a$.  Note that we need the range of $B$.  Then you have already obtained the following two inequalities in terms of these variables:
$$5C - 3 \le B \le 4C-1, \qquad B \ge Ce^{1/C}$$
The first inequality needs $5C-3 \le 4C-1 \implies C \le 2$.  Further $B > 0 \implies 4C-1 > 0 \implies C > \frac14$.  Thus $C$ can take only values in $(\frac14, 2]$.  Correspondingly, $B$ has to be in $(0, 7]$.
The second inequality further requires that we must have $\displaystyle B \ge \min_C Ce^{1/C}=  e$.  So it seems $B \in [e, 7]$ is the range.
