Question Number 188218 by pascal889 last updated on 26/Feb/23
Answered by cortano12 last updated on 27/Feb/23
![let log _x 10=(1/(log _(10) x)) =p , x>0 ,x≠1 ⇒2.log _x 10−(1/2)[log _x (((√x)/(100)))]−((((1/2)))/(1−log _x 10))=−(7/2) ⇒2p−(1/2)((1/2)−2p)−(1/(2−2p))=−(7/2) ⇒4p−(1/2)+2p−(1/(4−4p)) =−7 ⇒6p−(1/(4−4p))=−((13)/2) ⇒p=(((√(601))−1)/(24)) ⇒log _x (10)=(((√(601))−1)/(24)) ⇒(1/(log _(10) (x)))=(((√(601))−1)/( 24)) ⇒x=10^((24)/( (√(601))−1))](https://www.tinkutara.com/question/Q188249.png)
$$\:\mathrm{let}\:\mathrm{log}\:_{\mathrm{x}} \mathrm{10}=\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{10}} \mathrm{x}}\:=\mathrm{p}\:,\:\mathrm{x}>\mathrm{0}\:,\mathrm{x}\neq\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}.\mathrm{log}\:_{\mathrm{x}} \mathrm{10}−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{log}\:_{\mathrm{x}} \left(\frac{\sqrt{\mathrm{x}}}{\mathrm{100}}\right)\right]−\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{log}\:_{\mathrm{x}} \mathrm{10}}=−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2p}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2p}\right)−\frac{\mathrm{1}}{\mathrm{2}−\mathrm{2p}}=−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4p}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2p}−\frac{\mathrm{1}}{\mathrm{4}−\mathrm{4p}}\:=−\mathrm{7} \\ $$$$\Rightarrow\mathrm{6p}−\frac{\mathrm{1}}{\mathrm{4}−\mathrm{4p}}=−\frac{\mathrm{13}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{p}=\frac{\sqrt{\mathrm{601}}−\mathrm{1}}{\mathrm{24}}\: \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{10}\right)=\frac{\sqrt{\mathrm{601}}−\mathrm{1}}{\mathrm{24}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{x}\right)}=\frac{\sqrt{\mathrm{601}}−\mathrm{1}}{\:\mathrm{24}} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{10}^{\frac{\mathrm{24}}{\:\sqrt{\mathrm{601}}−\mathrm{1}}} \: \\ $$