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Question Number 147999 by iloveisrael last updated on 25/Jul/21

	Answered by Canebulok last updated on 25/Jul/21

	$S o l u t i o n :$ $\Rightarrow 16 \cdot {(2 x)}^{\frac{1}{(\frac{x}{2})}} = {(8 x)}^{\frac{1}{2}}$ $\Rightarrow 16 \cdot {(2 x)}^{\frac{2}{x}} = {(8 x)}^{\frac{1}{2}}$ $\Rightarrow 2^{4} \cdot {(2)}^{\frac{2}{x}} \cdot {(x)}^{\frac{2}{x}} = {(8 x)}^{\frac{1}{2}}$ $\Rightarrow {(2)}^{\frac{2}{x} + 4} \cdot {(x)}^{\frac{2}{x}} = {(8 x)}^{\frac{1}{2}}$ $\Rightarrow {(2)}^{4 x + 2} \cdot {(x)}^{2} = {(8 x)}^{\frac{x}{2}}$ $\Rightarrow {(2)}^{4 x + 2} \cdot {(x)}^{2} = {(8)}^{\frac{x}{2}} \cdot {(x)}^{\frac{x}{2}}$ $\Rightarrow {(2)}^{4 x + 2 - \frac{3 x}{2}} = {(x)}^{\frac{x}{2} - 2}$ $\Rightarrow {(2)}^{\frac{5 x}{2} + 2} = {(x)}^{\frac{x}{2} - 2}$ $\Rightarrow {(2)}^{5 x + 4} = {(x)}^{x - 4}$ $\Rightarrow 16 = {(x)}^{x - 4} \cdot {(2)}^{- 5 x}$ $\Rightarrow 16 = {(\frac{x}{32})}^{x} \cdot x^{- 4}$ $\Rightarrow 16 x^{4} = {(\frac{x}{32})}^{x}$ $\Rightarrow 2 x = {(\frac{x}{32})}^{\frac{x}{4}}$ $\Rightarrow^{8} \sqrt{2 x} = {(\frac{x}{32})}^{(\frac{x}{32})}$ $\Rightarrow \frac{l n (2 x)}{8} = l n (\frac{x}{32}) \cdot e^{l n (\frac{x}{32})}$ $\Rightarrow W (\frac{l n (2 x)}{8}) = l n (\frac{x}{32})$ $\Rightarrow \frac{W (\frac{l n (2 x)}{8})}{l n (\frac{2 x}{64})} = 1$ $\Rightarrow \frac{W (\frac{l n (2 x)}{8})}{(\frac{l n (2 x) - l n (64)}{8})} = 8$
	Commented by iloveisrael last updated on 25/Jul/21

	$If x = 8 \Rightarrow 16 . \sqrt[\frac{8}{2}]{16} = \sqrt{64}$ $\Rightarrow 16 . \sqrt[4]{16} = 8$ $\Rightarrow 16 \times 2 = 32 = 6 ? ? ?$
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