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Question-172264




Question Number 172264 by Mikenice last updated on 25/Jun/22
Answered by Rasheed.Sindhi last updated on 25/Jun/22
log_2 (log_8 x)=log_8 (log_2 x)      log_2 (((log_2 x )/(log_2 8 )))=((log_2 (log_2 x) )/(log_2 8))    log_2 (((log_2 x )/(3 )))=((log_2 (log_2 x) )/3)    3{log_2 (log_2 x)−log_2 3}=log_2 (log_2 x)  log_2 x=y  3log_2 y−3log_2 3=log_2 y  2log_2 y−3log_2 3=0    log_2 y^2 =log_2 3^3          y^2 =3^3       (log_2 x )^2 =27
$$\mathrm{log}_{\mathrm{2}} \left(\mathrm{log}_{\mathrm{8}} {x}\right)=\mathrm{log}_{\mathrm{8}} \left(\mathrm{log}_{\mathrm{2}} {x}\right)\:\:\:\: \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{log}_{\mathrm{2}} {x}\:}{\mathrm{log}_{\mathrm{2}} \mathrm{8}\:}\right)=\frac{\mathrm{log}_{\mathrm{2}} \left(\mathrm{log}_{\mathrm{2}} {x}\right)\:}{\mathrm{log}_{\mathrm{2}} \mathrm{8}}\:\: \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{log}_{\mathrm{2}} {x}\:}{\mathrm{3}\:}\right)=\frac{\mathrm{log}_{\mathrm{2}} \left(\mathrm{log}_{\mathrm{2}} {x}\right)\:}{\mathrm{3}}\:\: \\ $$$$\mathrm{3}\left\{\mathrm{log}_{\mathrm{2}} \left(\mathrm{log}_{\mathrm{2}} {x}\right)−\mathrm{log}_{\mathrm{2}} \mathrm{3}\right\}=\mathrm{log}_{\mathrm{2}} \left(\mathrm{log}_{\mathrm{2}} {x}\right) \\ $$$$\mathrm{log}_{\mathrm{2}} {x}={y} \\ $$$$\mathrm{3log}_{\mathrm{2}} {y}−\mathrm{3log}_{\mathrm{2}} \mathrm{3}=\mathrm{log}_{\mathrm{2}} {y} \\ $$$$\mathrm{2log}_{\mathrm{2}} {y}−\mathrm{3log}_{\mathrm{2}} \mathrm{3}=\mathrm{0} \\ $$$$\:\:\mathrm{log}_{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{log}_{\mathrm{2}} \mathrm{3}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:{y}^{\mathrm{2}} =\mathrm{3}^{\mathrm{3}} \\ $$$$\:\:\:\:\left(\mathrm{log}_{\mathrm{2}} {x}\:\right)^{\mathrm{2}} =\mathrm{27} \\ $$
Commented by Tawa11 last updated on 25/Jun/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 25/Jun/22
Thanks miss!
$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\mathrm{miss}! \\ $$
Answered by greougoury555 last updated on 25/Jun/22
 let log _8 x = u⇒log _2 x = 3u  ⇒ log _2 (u)=log _8 (3u)  ⇒log _2 (u^3 )=log _2 (3u)  ⇒u^3 −3u=0 ⇒u(u^2 −3)=0  ⇒ { ((u=0⇒x=1)),(((log _8 x)^2 =3⇒(log_2 x)^2 = 27 )) :}
$$\:{let}\:\mathrm{log}\:_{\mathrm{8}} {x}\:=\:{u}\Rightarrow\mathrm{log}\:_{\mathrm{2}} {x}\:=\:\mathrm{3}{u} \\ $$$$\Rightarrow\:\mathrm{log}\:_{\mathrm{2}} \left({u}\right)=\mathrm{log}\:_{\mathrm{8}} \left(\mathrm{3}{u}\right) \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{2}} \left({u}^{\mathrm{3}} \right)=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}{u}\right) \\ $$$$\Rightarrow{u}^{\mathrm{3}} −\mathrm{3}{u}=\mathrm{0}\:\Rightarrow{u}\left({u}^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{{u}=\mathrm{0}\Rightarrow{x}=\mathrm{1}}\\{\left(\mathrm{log}\:_{\mathrm{8}} {x}\right)^{\mathrm{2}} =\mathrm{3}\Rightarrow\left(\mathrm{log}_{\mathrm{2}} {x}\right)^{\mathrm{2}} =\:\mathrm{27}\:}\end{cases} \\ $$

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