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.
 A: The logarithms will be rational only if $a$ is of the form $2^b$ where $b$ is an integer
$$f(b)=\dfrac{5b}2+\dfrac{96}b-2^{b-3}$$
Now using AM-GM inequality, the Left hand side $$\ge\sqrt{240}>15$$
$2^{b-3}\ge15\implies b\ge7$
By trial, $b=8$ is a solution
A: An algebraic solution
Let $x=4+t$. Then $128+40t+5t^2=(128+32t)2^{2t}$.
First suppose $t\ge0$. Then $2^{2t}\ge{1+t}$ and so $$128+40t+5t^2\ge128+160t+32t^2.$$Then $120t+27t^2\le0$ and so $t=0$.
Next suppose that $t\le0$ and let $s=-t$. Then $$2^{2s}(128-40s+5s^2)=128-32s$$$$128-32s\ge(1+s)(128-40s+5s^2)$$Therefore $0\ge5s(s^2-7s+24)$ and $s=0$.
The only solution is $t=0$ i.e. $x=4$.
A: 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 $$0<x<4$$.
