Non-graphical solution to $5\log_{4}a\ + 48\log_{a}4 = \frac{a}{8}$ It is required to solve

$$5 \log_{4}a\ + 48\log_{a}4 = \frac{a}{8}$$

Here is my attempt, 
Let $$ x = \log_{4}a$$, then $$a = 2^{2x}$$
And our equation becomes
$$ 5x^{2} - x\cdot 2^{2x-3} + 48 = 0$$
But this is as far as I can go. I've tried several substitutions but no progress.

To localize the solution x=4 you can write
$$
5x^2  - x \cdot 2^{2x - 3}  + 48 = 0
$$
as
$$
5x + \frac{{48}}
{x} = 2^{2x - 3}
$$
This tell you that x must be positive.  Now, first search for integer solutions. It is obvious that $$x=1$$ is not a solution and therefore you can imagine that $$x \geq 2$$. In this case the RHS is integer and therefore x must be a divisor of 48. Among these numbers you have that only x=4 and x=12 let the LHS a power of 2. Since x=12 is not a solution while x=4 does you have that the only integer solution is x=4.
Now let be
$$
f(x) = 2^{2x - 3}  - 5x
$$
and 
$$g(x)=\frac{48}{x}$$
It is g(x) <12 if x>4 and g(x)>12 if $$0<x<4$$. We will prove that f(x)>12 if x>4 and f(x)<12 per $$0<x<4$$. Namely let be x=4+t with t>0. It is
$$
\begin{gathered}
  2^{2(4 + t) - 3}  - 5(4 + t) > 12 \Leftrightarrow  \hfill \\
  2^{5 + 2t}  - 32 - 5t > 0 \Leftrightarrow  \hfill \\
  32 \cdot 4^t  - 32 - 5t > 0 \hfill \\ 
\end{gathered} 
$$
But with t>0 it is
$$4^t>1+t$$ 
therefore
$$32 \cdot 4^t  - 32 - 5t > 32\left( {1 + t} \right) - 32 - 5t = 27t > 0
$$
Hence there is no solutions if x>4. If $$0<x<4$$ we can write $$x=4-t$$ with $$0<t<4$$. It is
$$
\begin{gathered}
  2^{2(4 - t) - 3}  - 5(4 - t) < 12 \Leftrightarrow  \hfill \\
  2^{5 - 2t}  - 32 + 5t < 0 \Leftrightarrow  \hfill \\
  32 \cdot \frac{1}
{{4^t }} - 32 + 5t < 0 \hfill \\ 
\end{gathered} 
$$
We have that
$$
\begin{gathered}
  32 \cdot \frac{1}
{{4^t }} - 32 + 5t < 32 \cdot \frac{1}
{{1 + t}} - 32 + 5t =  \hfill \\
   = \frac{{ - 32t}}
{{1 + t}} + 5t = \frac{{ - 27t + 5t^2 }}
{{1 + t}} = t\frac{{5t - 27}}
{{1 + t}} < 0 \hfill \\ 
\end{gathered} 
$$
which is negative for $$0<t<4$$.
Therefore there are no solutions with $$0<x<4$$.
