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$\begin{array}{|rcll|} \hline 8^{ \sqrt{4x+2} } > \dfrac{512^{2x}} { 8^{ \sqrt{16x+8} } } \\ 8^{ \sqrt{4x+2} } > \dfrac{512^{2x}} { 8^{ 2\cdot \sqrt{4x+2} } } \quad & | \quad512 = 8^3 \\ 8^{ \sqrt{4x+2} } > \dfrac{8^{3\cdot 2x}} { 8^{ 2\cdot \sqrt{4x+2} } } \\ 8^{ \sqrt{4x+2} } > \dfrac{8^{6x}} { 8^{ 2\cdot \sqrt{4x+2} } } \\ 8^{ \sqrt{4x+2} } > 8^{6x-2\cdot \sqrt{4x+2} } \\ \sqrt{4x+2} > 6x-2\cdot \sqrt{4x+2} \\ 3\cdot \sqrt{4x+2} > 6x \quad & | \quad : 3\\ \sqrt{4x+2} > 2x \\ \hline \end{array}$

Let's find the domain:

$\begin{array}{|rcll|} \hline 4x+2 &\ge & 0 \quad & | \quad -2\\ 4x &\ge & -2 \quad & | \quad : 4 \\ x &\ge & -\frac12 \\ \hline \end{array}$

The domain is $[-\frac12, \infty)$

case differentiation

1. $x\ge 0$

$\begin{array}{|rcll|} \hline \sqrt{4x+2} &>& 2x \quad & | \quad \text{square both sides}\\ 4x+2 &>& 4x^2 \quad & | \quad -4x \\ 2 &>& 4x^2-4x \quad & | \quad :4 \\ \frac12 &>& x^2-x \\ \frac12 &>& \left(x-\frac12 \right)^2-\frac14 \quad & | \quad + \frac14\\ \frac12+ \frac14 &>& \left(x-\frac12 \right)^2 \\ \frac34 &>& \left(x-\frac12 \right)^2 \quad & | \quad \sqrt{ } \\ \frac{\sqrt{3}}{2} &>& | x-\frac12 | \\ | x-\frac12 | &<& \frac{\sqrt{3}}{2} \\ \Rightarrow \quad - \frac{\sqrt{3}}{2} &<& x-\frac12 < \frac{\sqrt{3}}{2} \quad & | \quad +\frac12 \\ - \frac{\sqrt{3}}{2} +\frac12 &<& x-\frac12 +\frac12 < \frac{\sqrt{3}}{2} +\frac12 \\ \frac{1-\sqrt{3}}{2} &<& x < \frac{1+\sqrt{3}}{2} \\ \hline \end{array}$

case differentiation
2. x < 0 , because the domain is $x \ge -\frac12$

$\begin{array}{|rcll|} \hline x &<& 0 \\ \text{the domain is } x &\ge& -\frac12 \text{ or } \\ -\frac12 & \le& x \\ \Rightarrow \quad -\frac12 & \le & x < 0 \\\\ \underbrace{\sqrt{4x+2}}_{\ge 0} &>& \underbrace{2x}_{<0} \quad & | \quad \text{always true}\\ \hline \end{array}$

together:

$\begin{array}{|rcll|} \hline \frac{1-\sqrt{3}}{2} &<& x < \frac{1+\sqrt{3}}{2} \\ -\frac12 & \le & x < 0 \\\\ \mathbf{ -\frac12 } & \mathbf{ \le } & \mathbf{ x < \frac {1+\sqrt{3}}{2} } \\ \hline \end{array}$

8^{√[4x + 2]} > 512^2x / 8^{√[16x + 8]}

8^{√[4x + 2]} > 512^2x / 8^{√[4(4x + 2)]} 

8^{√[4x + 2]} > 512^2x / 8 ^{2√[(4x + 2)]} 

8^{√[4x + 2]}  >  (8³)^2x / 8 ^{2√[(4x + 2)]} 

8^{√[4x + 2]}  > 8 ^{6x -  2√[(4x + 2)]}    we can solve for the exponents.......

 √[(4x + 2)]  > 6x -  2√[(4x + 2)]      

 √[(4x + 2)]  > 6x -  2√[(4x + 2)]     add  2√[(4x + 2)]  to both sides

3 √[(4x + 2)] > 6x      divide both sides by 3

√[(4x + 2)]   >  2x      square both sides

4x + 2  > 4x^2     divide through by 2

2x + 1  > 2x^2   

0 > 2x^2 - 2x - 1  

And the  solution  to this is     1/2 (1-sqrt(3)) < x <1/2 (1+sqrt(3))  

However, for the original equation √[(4x + 2)]   >  2x  ......this is  true if   -1/2 ≤  x <1/2 (1+sqrt(3))

But, since 1/2 (1-sqrt(3))  > -1/2 , we can expand the solution interval to    -1/2 ≤ x < 1/2 (1+sqrt(3))