Find solutions for $\lambda$ in this (diophantine?) equation? I am interested in finding the solutions to the equation below:
$$\lambda = \frac{2^c+3b}{2^d-3^a}$$
$$\lambda, a, b, c, d \in \Bbb Z^+$$
$$d>a,c$$
Edit:
My original version was actually:
$$\frac{3^a(\lambda) + 3(3b+2^c)+2^d}{2^e}= \lambda$$
$$e>a,c,d$$
Furthermore, when looking at this original equation, it became apparent that this was actually:
$$\frac{3^a \lambda+3^a b +\sum_{i=o}^{a-1} 3^i 2^{j_i}}{2^c}=\lambda$$
Where $j_i \in \Bbb Z^+ < d-1$.
At this point, I am lost as to how I should attempt to solve this.
Are there solutions to the original equations, other than $1$ and $2$? 
Can anyone find large solutions? ($a,b,c,d > 100$)?
I am only really interested in the large solutions. 
Also, is there a technique to find solutions to this as I am really not sure how to go about this.
 A: I would recommend to describe the original problem before the generalisation, because in its current form without any restrictions the solution is straightforward.
The equation in positive integers $a,b,c,d,\lambda$
$$
\lambda = \frac{2^c + 3 b}{2^d - 3^a}
$$
can immediately be rewritten as
$$
3 b = \lambda 2^d - 2^c - \lambda 3^a
$$
So the only constraint on the integers (using $d>c$) we find is that $3 | \lambda 2^d - 2^c = 2^c (\lambda 2^{d-c} - 1)$. This results in either a solution with $\lambda \equiv 1~\text{mod}~3$ and $d-c \equiv 0~\text{mod}~2$ or $\lambda \equiv 2~\text{mod}~3$ and $d-c \equiv 1~\text{mod}~2$. There are no solutions with $\lambda$ being a multiple of 3.
For any choice of positive $a$,$c$, and $\lambda$ (not a multiple of 3) either all even or odd values of $d > \max\left(c,\frac{\log 3^a + 2^c/\lambda}{\log 2}\right)$ will be enough to guarantee an integer solution for $b$. Note that the constraint $d>a$ is automatically satisfied.
The solution for the original problem is more or less the same. Rewriting it gives that
$$
9 b = \lambda 2^e - 2^d - \lambda 3^a - 3 \times 2^c
$$
Assuming $a,c,d,e$ are positive the RHS needs to be a multiple of 3, hence $\lambda \equiv 1(\text{or } 2)\text{ mod }3$ and $e - d \equiv 0 (\text{or }1) \text{ mod } 2$.
It gets a little bit more complicated, because the RHS needs to be a multiple of 9. This means that for each of the 4 terms you need to determine what its remainder modulo 9 will be and how that is related to the values of $c,d,e$ modulo 6 whereas for $\lambda$ it is modulo 9. Also you need to distinguish the cases $a=1$ and $a \geq 2$. After this analysis, which is not too difficult but just many cases, you can choose any suitable combination of $a,c,d,e,\lambda$ with the only additional restriction that $e$ should be large enough to make $b$ positive.   
