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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] > 5122x / 8√[16x + 8]
8√[4x + 2] > 5122x / 8√[4(4x + 2)]
8√[4x + 2] > 5122x / 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))