Standard question in log

Let $u = log_{4}(log_{3}x)$ and let us rewrite the above equation as:

$3u(2 \cdot \frac{u}{log_{4}(64)} - 1) \Rightarrow 3u(\frac{2}{3}u - 1) = -1;$

or $2u^2 - 3u + 1 = 0$ ;

or $(2u-1)(u-1) = 0;$

or $u = \frac{1}{2}, 1.$

Now we substitute back to obtain: $log_{4}(log_{3}x) = \frac{1}{2}, 1 \Rightarrow log_{3}x = 2, 4 \Rightarrow x = 9, 81.$ Hence, the sum of all such $x$ equals $9+ 81 = \boxed{90}.$

Let $\log_3 x = 4^{3y}$ . Then by substitution:

$3 \log_4 (\log_3 x) (2 \log_{64} (\log_3 x) - 1) = -1$

$3 \log_4 4^{3y} (2 \log_{64} 4^{3y} - 1) = -1$

$3 \cdot 3y \cdot (2 \cdot y - 1) = -1$

$18y^2 - 9y + 1 = 0$

$(3y - 1)(6y - 1) = 0$

$3y = 1$ or $6y = 1$

If $3y = 1$ , then:

$\log_3 x = 4^{3y}$

$\log_3 x = 4^1$

$x = 3^4 = 81$

If $6y = 1$ , then $3y = \frac{1}{2}$ and:

$\log_3 x = 4^{3y}$

$\log_3 x = 4^{\frac{1}{2}}$

$x = 3^2 = 9$

Therefore, the sum of the two solutions for $x$ is $81 + 9 = \boxed{90}$ .