Question Number 127519 by O Predador last updated on 30/Dec/20
Answered by mahdipoor last updated on 30/Dec/20
![0.0833...=0.083^− =0.08+0.003^− = (8/(100))+(1/(300))=((25)/(300))=(1/(12))=((1×2)/(1×2×3×4))=((2!)/(4!))=(([log_2 (x^2 )−2]!)/([log_2 (x)+2]!)) ⇒x=4](Q127522.png)
$$\mathrm{0}.\mathrm{0833}...=\mathrm{0}.\mathrm{08}\overset{−} {\mathrm{3}}=\mathrm{0}.\mathrm{08}+\mathrm{0}.\mathrm{00}\overset{−} {\mathrm{3}}= \\ $$$$\frac{\mathrm{8}}{\mathrm{100}}+\frac{\mathrm{1}}{\mathrm{300}}=\frac{\mathrm{25}}{\mathrm{300}}=\frac{\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{1}×\mathrm{2}}{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}}=\frac{\mathrm{2}!}{\mathrm{4}!}=\frac{\left[{log}_{\mathrm{2}} \left({x}^{\mathrm{2}} \right)−\mathrm{2}\right]!}{\left[{log}_{\mathrm{2}} \left({x}\right)+\mathrm{2}\right]!} \\ $$$$\Rightarrow{x}=\mathrm{4} \\ $$