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

Question

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.

Answer 1

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 $$0<x<4$$.

Answer 2

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

Answer 3

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. $x=4$.