Question Number 175040 by andres_chu last updated on 17/Aug/22
$$ \\ $$hello, please, someone help me to correct the equation? It's typed wrong and I can't find where
$${Solve}: \\ $$$$\mathrm{5},\mathrm{76}\left[\frac{\mathrm{log}_{{a}} \left(\sqrt{\mathrm{log}\:_{{b}} \left(\sqrt{{a}}\right)}\right)}{\mathrm{log}\left(\sqrt{\mathrm{log}\left({a}\right)}\right)}\:+\:\mathrm{log}_{\mathrm{log}\:\left({a}\right)} \left(\mathrm{2}\right)\right]\sqrt[{\mathrm{5}}]{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)}\:+\:\frac{\mathrm{log}_{\mathrm{2}} \left({x}\right)}{\mathrm{25}}\:=\:\left[\mathrm{log}\:_{\mathrm{2}} \left({x}\right)\right]^{\frac{\mathrm{3}}{\mathrm{5}}} \\ $$$${Answers} \\ $$$${x}_{\mathrm{1}} =\mathrm{1}\:,\:{x}_{\mathrm{2}} =\mathrm{2}^{\mathrm{243}} \:,\:{x}_{\mathrm{3}} =\mathrm{2}^{−\mathrm{243}} \:,\:{x}_{\mathrm{4}} =\mathrm{2}^{\mathrm{1024}} \:,\:{x}_{\mathrm{5}} =\mathrm{2}^{−\mathrm{1024}} \\ $$
$${Solve}: \\ $$$$\mathrm{5},\mathrm{76}\left[\frac{\mathrm{log}_{{a}} \left(\sqrt{\mathrm{log}\:_{{b}} \left(\sqrt{{a}}\right)}\right)}{\mathrm{log}\left(\sqrt{\mathrm{log}\left({a}\right)}\right)}\:+\:\mathrm{log}_{\mathrm{log}\:\left({a}\right)} \left(\mathrm{2}\right)\right]\sqrt[{\mathrm{5}}]{\mathrm{log}\:_{\mathrm{2}} \left({x}\right)}\:+\:\frac{\mathrm{log}_{\mathrm{2}} \left({x}\right)}{\mathrm{25}}\:=\:\left[\mathrm{log}\:_{\mathrm{2}} \left({x}\right)\right]^{\frac{\mathrm{3}}{\mathrm{5}}} \\ $$$${Answers} \\ $$$${x}_{\mathrm{1}} =\mathrm{1}\:,\:{x}_{\mathrm{2}} =\mathrm{2}^{\mathrm{243}} \:,\:{x}_{\mathrm{3}} =\mathrm{2}^{−\mathrm{243}} \:,\:{x}_{\mathrm{4}} =\mathrm{2}^{\mathrm{1024}} \:,\:{x}_{\mathrm{5}} =\mathrm{2}^{−\mathrm{1024}} \\ $$
Answered by a.lgnaoui last updated on 20/Aug/22
$$ \\ $$$${the}\:{equation}\:{will}\:{be}: \\ $$$$\mathrm{5},\mathrm{76}\left[\frac{\mathrm{log}\left(\sqrt{\mathrm{log}\left(\sqrt{\left(\mathrm{a}\right)}\:\right.}\right.}{\mathrm{log}\left(\sqrt{\mathrm{log}\left(\mathrm{a}\right)}\right.}\:+\mathrm{log}_{\mathrm{log}\left(\mathrm{a}\right)} \left(\mathrm{2}\right)\right]\left(\mathrm{log}\left(\mathrm{x}\right)_{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{5}} +\frac{\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{25}}=\left(\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)\right)^{\mathrm{3}/\mathrm{5}} \\ $$$$\:\:\frac{\mathrm{log}\left(\sqrt{\mathrm{log}\left(\sqrt{\left.\mathrm{a}\left.\right)\right)}\right.}\right.}{\mathrm{log}\left(\sqrt{\mathrm{log}\left(\mathrm{a}\right)}\right.}\:+\mathrm{log}_{\mathrm{log}\left(\mathrm{a}\right)} \left(\mathrm{2}\right)\:=\:\left[\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{a}\right)\right)}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{log}\left(\mathrm{a}\right)\right)}\:\:+\frac{\mathrm{log}\left(\mathrm{2}\right)}{\mathrm{log}\left(\mathrm{log}\left(\mathrm{a}\right)\right)}\right] \\ $$$$\frac{\mathrm{log}\left(\mathrm{log}\left(\mathrm{a}\right)\right)−\mathrm{log}\left(\mathrm{2}\right)}{\mathrm{log}\left(\mathrm{log}\left(\mathrm{a}\right)\right)}+\frac{\mathrm{log}\left(\mathrm{2}\right)}{\mathrm{log}\left(\mathrm{log}\left(\mathrm{a}\right)\right)}=\mathrm{1} \\ $$$$\mathrm{5},\mathrm{76}\left(\mathrm{log}_{\mathrm{2}} \mathrm{x}\right)^{\mathrm{1}/\mathrm{5}} +\frac{\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{25}}=\left(\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)\right)^{\mathrm{3}/\mathrm{5}} \\ $$$$\mathrm{posons}\:\left(\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)\right)^{\mathrm{1}/\mathrm{5}} =\mathrm{X} \\ $$$$\mathrm{X}\left[\mathrm{5},\mathrm{76}+\frac{\mathrm{X}^{\mathrm{4}} }{\mathrm{25}}−\mathrm{X}^{\mathrm{2}} =\mathrm{0}\right]\Rightarrow\mathrm{X}^{\mathrm{4}} −\mathrm{25X}^{\mathrm{2}} +\mathrm{144}=\mathrm{0} \\ $$$$\mathrm{Z}=\mathrm{X}^{\mathrm{2}} \:\:\:\:\mathrm{Z}^{\mathrm{2}} −\mathrm{25Z}+\mathrm{144}=\mathrm{0}\:\:\:\Delta=\mathrm{7}^{\mathrm{2}} \:\:\mathrm{Z}=\frac{\mathrm{25}\pm\mathrm{7}}{\mathrm{2}}=\left(\mathrm{9},\mathrm{16}\right)\:\:\mathrm{X}=\left(\mathrm{3},\mathrm{4}\right) \\ $$$$\left[\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)\right]^{\mathrm{1}/\mathrm{5}} =\mathrm{3}\:\:\:\:\Rightarrow\mathrm{log}\left(\mathrm{x}\right)\:=\mathrm{3}^{\mathrm{5}} \mathrm{log}\left(\mathrm{2}\right)^{\mathrm{5}} \:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{7},\mathrm{68}×\mathrm{10}^{\mathrm{16}} \:\: \\ $$$$\mathrm{x}=\mathrm{4}\:\:\:\:\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{4}^{\mathrm{5}} \:\:\:\:\:\:\left[\mathrm{log}\left(\mathrm{x}\right)=\mathrm{4}^{\mathrm{5}} \mathrm{log}\left(\mathrm{2}\right)^{\mathrm{5}} \:\:\:\mathrm{x}=\mathrm{1},\mathrm{428}×\mathrm{10}^{\mathrm{71}} \right. \\ $$