Question Number 183777 by Michaelfaraday last updated on 30/Dec/22
Commented by Frix last updated on 30/Dec/22
$${k}=\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$
Answered by HeferH last updated on 30/Dec/22
$$\mathrm{24}^{\mathrm{1}\:−\:\frac{\mathrm{1}}{{k}}} \:=\:\mathrm{3} \\ $$$$\:\frac{\mathrm{24}}{\mathrm{24}^{\frac{\mathrm{1}}{{k}}} }\:=\:\mathrm{3} \\ $$$$\:\mathrm{24}\:=\:\mathrm{3}\centerdot\mathrm{24}^{\frac{\mathrm{1}}{{k}}} \\ $$$$\:\mathrm{8}\:=\:\mathrm{24}^{\frac{\mathrm{1}}{{k}}} \\ $$$$\:\mathrm{2}^{\mathrm{3}{k}} =\:\mathrm{24} \\ $$$$\:\mathrm{2}^{{k}} =\:\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$
Commented by Michaelfaraday last updated on 30/Dec/22
$${thanks}\:{sir} \\ $$
Answered by cortano1 last updated on 30/Dec/22
$$\:\mathrm{1}−\frac{\mathrm{1}}{{k}}\:=\:\mathrm{log}\:_{\mathrm{24}} \left(\mathrm{3}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{k}}\:=\:\mathrm{1}−\mathrm{log}\:_{\mathrm{24}} \left(\mathrm{3}\right)=\mathrm{log}\:_{\mathrm{24}} \left(\mathrm{8}\right) \\ $$$$\Rightarrow{k}=\mathrm{log}\:_{\mathrm{8}} \left(\mathrm{24}\right)=\frac{\mathrm{1}}{\mathrm{3}}\left\{\mathrm{3}+\mathrm{log}\:_{\mathrm{2}} \mathrm{3}\right\} \\ $$$$\therefore\:\mathrm{2}^{{k}} \:=\:\mathrm{2}^{\mathrm{1}+\mathrm{log}\:_{\mathrm{2}} \left(\left(\sqrt[{\mathrm{3}}]{\mathrm{3}}\right)\right.} =\:\mathrm{2}.\sqrt[{\mathrm{3}}]{\mathrm{3}}\: \\ $$$$ \\ $$$$ \\ $$
Commented by Michaelfaraday last updated on 30/Dec/22
$${thanks}\:{sir} \\ $$