Question Number 130168 by Adel last updated on 23/Jan/21
$$\mathrm{log}\underset{\mathrm{X}} {\mathrm{2}}\centerdot\mathrm{log}\:\underset{\mathrm{2X}} {\mathrm{2}}=\mathrm{log}\:\underset{\mathrm{4X}} {\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{X}=? \\ $$
Commented by liberty last updated on 23/Jan/21
$$\mathrm{do}\:\mathrm{you}\:\mathrm{meant}\:\mathrm{log}\:_{{x}} \left(\mathrm{2}\right).\mathrm{log}\:_{\mathrm{2}{x}} \left(\mathrm{2}\right)\:=\:\mathrm{log}\:_{\mathrm{4}{x}} \left(\mathrm{2}\right)? \\ $$
Answered by liberty last updated on 23/Jan/21
![(1/(log _2 (x))).(1/(log _2 (2x))) = (1/(log _2 (4x))) 2+log _2 (x)=log _2 (x)[ 1+log _2 (x) ] let log _2 (x)=p ⇒2+p = p+p^2 ; p^2 =2 → { ((p=(√2))),((p=−(√2))) :} ⇒log _2 (x)=± (√2) ⇒x = 2^(± (√2)) ; x = { (2^(√2) ),((1/2^(√2) )) :}](https://www.tinkutara.com/question/Q130171.png)
$$\:\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right)}.\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{2x}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{4x}\right)} \\ $$$$\:\mathrm{2}+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right)\left[\:\mathrm{1}+\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right)\:\right] \\ $$$$\mathrm{let}\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{p} \\ $$$$\Rightarrow\mathrm{2}+\mathrm{p}\:=\:\mathrm{p}+\mathrm{p}^{\mathrm{2}} \:;\:\mathrm{p}^{\mathrm{2}} =\mathrm{2}\:\rightarrow\begin{cases}{\mathrm{p}=\sqrt{\mathrm{2}}}\\{\mathrm{p}=−\sqrt{\mathrm{2}}}\end{cases} \\ $$$$\:\Rightarrow\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{x}\right)=\pm\:\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}\:=\:\mathrm{2}^{\pm\:\sqrt{\mathrm{2}}} \:;\:\mathrm{x}\:=\:\begin{cases}{\mathrm{2}^{\sqrt{\mathrm{2}}} }\\{\frac{\mathrm{1}}{\mathrm{2}^{\sqrt{\mathrm{2}}} }}\end{cases}\: \\ $$
Commented by Adel last updated on 23/Jan/21
$$\mathrm{thanks}\:\mathrm{sir} \\ $$
Answered by MJS_new last updated on 23/Jan/21
$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:{x}}×\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{2}{x}}=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\mathrm{4}{x}} \\ $$$${t}=\mathrm{ln}\:{x} \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{{t}}×\frac{\mathrm{ln}\:\mathrm{2}}{{t}+\mathrm{ln}\:\mathrm{2}}=\frac{\mathrm{ln}\:\mathrm{2}}{{t}+\mathrm{2ln}\:\mathrm{2}} \\ $$$${t}\left({t}+\mathrm{ln}\:\mathrm{2}\right)=\mathrm{ln}\:\mathrm{2}\:\left({t}+\mathrm{2ln}\:\mathrm{2}\right) \\ $$$${t}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{ln}\:\mathrm{2}\right)^{\mathrm{2}} \\ $$$${t}=\pm\sqrt{\mathrm{2}}\mathrm{ln}\:\mathrm{2} \\ $$$${x}=\mathrm{2}^{−\sqrt{\mathrm{2}}} \vee{x}=\mathrm{2}^{\sqrt{\mathrm{2}}} \\ $$
Commented by liberty last updated on 23/Jan/21
$$\mathrm{yes}…\mathrm{my}\:\mathrm{answer}\:\mathrm{correct}.. \\ $$
Commented by Adel last updated on 23/Jan/21
$$\mathrm{tanks}\:\mathrm{sir} \\ $$