Question Number 147999 by iloveisrael last updated on 25/Jul/21
Answered by Canebulok last updated on 25/Jul/21
$$\: \\ $$$$\boldsymbol{{Solution}}: \\ $$$$\Rightarrow\:\mathrm{16}\centerdot\left(\mathrm{2}{x}\right)^{\frac{\mathrm{1}}{\left(\frac{{x}}{\mathrm{2}}\right)}} \:=\:\left(\mathrm{8}{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\: \\ $$$$\Rightarrow\:\mathrm{16}\centerdot\left(\mathrm{2}{x}\right)^{\frac{\mathrm{2}}{{x}}} \:=\:\left(\mathrm{8}{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\: \\ $$$$\Rightarrow\:\mathrm{2}^{\mathrm{4}} \centerdot\left(\mathrm{2}\right)^{\frac{\mathrm{2}}{{x}}} \centerdot\left({x}\right)^{\frac{\mathrm{2}}{{x}}} \:=\:\left(\mathrm{8}{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\: \\ $$$$\Rightarrow\:\left(\mathrm{2}\right)^{\frac{\mathrm{2}}{{x}}+\mathrm{4}} \centerdot\left({x}\right)^{\frac{\mathrm{2}}{{x}}} \:=\:\left(\mathrm{8}{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\: \\ $$$$\Rightarrow\:\left(\mathrm{2}\right)^{\mathrm{4}{x}+\mathrm{2}} \centerdot\left({x}\right)^{\mathrm{2}} \:=\:\left(\mathrm{8}{x}\right)^{\frac{{x}}{\mathrm{2}}} \\ $$$$\: \\ $$$$\Rightarrow\:\left(\mathrm{2}\right)^{\mathrm{4}{x}+\mathrm{2}} \centerdot\left({x}\right)^{\mathrm{2}} \:=\:\left(\mathrm{8}\right)^{\frac{{x}}{\mathrm{2}}} \centerdot\left({x}\right)^{\frac{{x}}{\mathrm{2}}} \\ $$$$\: \\ $$$$\Rightarrow\:\left(\mathrm{2}\right)^{\mathrm{4}{x}+\mathrm{2}−\frac{\mathrm{3}{x}}{\mathrm{2}}} \:=\:\left({x}\right)^{\frac{{x}}{\mathrm{2}}−\mathrm{2}} \\ $$$$\: \\ $$$$\Rightarrow\:\left(\mathrm{2}\right)^{\frac{\mathrm{5}{x}}{\mathrm{2}}+\mathrm{2}} \:=\:\left({x}\right)^{\frac{{x}}{\mathrm{2}}−\mathrm{2}} \\ $$$$\: \\ $$$$\Rightarrow\:\left(\mathrm{2}\right)^{\mathrm{5}{x}+\mathrm{4}} \:=\:\left({x}\right)^{{x}−\mathrm{4}} \\ $$$$\: \\ $$$$\Rightarrow\:\mathrm{16}\:=\:\left({x}\right)^{{x}−\mathrm{4}} \centerdot\left(\mathrm{2}\right)^{−\mathrm{5}{x}} \\ $$$$\: \\ $$$$\Rightarrow\:\mathrm{16}\:=\:\left(\frac{{x}}{\mathrm{32}}\right)^{{x}} \centerdot{x}^{−\mathrm{4}} \\ $$$$\: \\ $$$$\Rightarrow\:\mathrm{16}{x}^{\mathrm{4}} \:=\:\left(\frac{{x}}{\mathrm{32}}\right)^{{x}} \\ $$$$\: \\ $$$$\Rightarrow\:\mathrm{2}{x}\:=\:\left(\frac{{x}}{\mathrm{32}}\right)^{\frac{{x}}{\mathrm{4}}} \\ $$$$\: \\ $$$$\Rightarrow\:^{\mathrm{8}} \sqrt{\mathrm{2}{x}}\:=\:\left(\frac{{x}}{\mathrm{32}}\right)^{\left(\frac{{x}}{\mathrm{32}}\right)} \\ $$$$\: \\ $$$$\Rightarrow\:\frac{{ln}\left(\mathrm{2}{x}\right)}{\mathrm{8}}\:=\:{ln}\left(\frac{{x}}{\mathrm{32}}\right)\centerdot{e}^{{ln}\left(\frac{{x}}{\mathrm{32}}\right)} \\ $$$$\: \\ $$$$\Rightarrow\:{W}\left(\frac{{ln}\left(\mathrm{2}{x}\right)}{\mathrm{8}}\right)\:=\:{ln}\left(\frac{{x}}{\mathrm{32}}\right) \\ $$$$\: \\ $$$$\Rightarrow\:\frac{{W}\left(\frac{{ln}\left(\mathrm{2}{x}\right)}{\mathrm{8}}\right)}{{ln}\left(\frac{\mathrm{2}{x}}{\mathrm{64}}\right)}\:=\:\mathrm{1} \\ $$$$\: \\ $$$$\Rightarrow\:\frac{{W}\left(\frac{{ln}\left(\mathrm{2}{x}\right)}{\mathrm{8}}\right)}{\left(\frac{{ln}\left(\mathrm{2}{x}\right)−{ln}\left(\mathrm{64}\right)}{\mathrm{8}}\right)}\:=\:\mathrm{8} \\ $$$$\: \\ $$
Commented by iloveisrael last updated on 25/Jul/21
$$\mathrm{If}\:\mathrm{x}=\mathrm{8}\:\Rightarrow\mathrm{16}.\sqrt[{\frac{\mathrm{8}}{\mathrm{2}}}]{\mathrm{16}}\:=\:\sqrt{\mathrm{64}} \\ $$$$\Rightarrow\mathrm{16}.\sqrt[{\mathrm{4}}]{\mathrm{16}}\:=\:\mathrm{8} \\ $$$$\Rightarrow\:\mathrm{16}×\mathrm{2}\:=\:\mathrm{32}\:=\:\mathrm{6}\:???\: \\ $$