Question Number 86042 by M±th+et£s last updated on 26/Mar/20
$${solve}:\:\:\lfloor\:\sqrt{{x}}\:\rfloor=\lfloor\frac{{x}}{\mathrm{2}}\rfloor \\ $$
Answered by Rio Michael last updated on 26/Mar/20
$$\:\lfloor\sqrt{{x}}\:\rfloor\:=\:\lfloor\frac{{x}}{\mathrm{2}}\rfloor \\ $$$$\:\Rightarrow\:\sqrt{{x}}\:=\:\frac{{x}}{\mathrm{2}} \\ $$$$\:\:\:{x}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:{x}^{\mathrm{2}} −\mathrm{4}{x}\:=\:\mathrm{0}\:\Leftrightarrow\:{x}\:=\:\mathrm{0}\:{or}\:{x}\:=\:\mathrm{4} \\ $$
Commented by M±th+et£s last updated on 26/Mar/20
$$\mathrm{2}^{{log}_{\mathrm{2}} \left({x}−\mathrm{2}\right)} =\mathrm{2}^{{log}_{\mathrm{2}} \left(\mathrm{2}{x}−\mathrm{3}\right)} \\ $$$${x}−\mathrm{2}=\mathrm{2}{x}−\mathrm{3} \\ $$$$−{x}=−\mathrm{1} \\ $$$${x}=\mathrm{1}\:\:\:\:\:\:\:{log}_{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}\right)={log}_{\mathrm{2}} \left(\mathrm{2}{x}−\mathrm{3}\right) \\ $$$${x}=\varnothing \\ $$
Commented by Rio Michael last updated on 26/Mar/20
$$\mathrm{s}{ame}\:{process} \\ $$
Commented by M±th+et£s last updated on 26/Mar/20
$${ok}\:{thank}\:{you}\:{sir} \\ $$
Commented by Rio Michael last updated on 26/Mar/20
$$\mathrm{how}\:\mathrm{would}\:\mathrm{you}\:\mathrm{solve} \\ $$$$\:\mathrm{log}_{\mathrm{2}} \:\left({x}−\mathrm{2}\right)\:=\:\mathrm{log}_{\mathrm{2}} \left(\:\mathrm{2}{x}−\mathrm{3}\right)? \\ $$
Commented by M±th+et£s last updated on 26/Mar/20
$${sir}\:{why}\:\lfloor\sqrt{{x}}\rfloor=\lfloor\frac{{x}}{\mathrm{2}}\rfloor \\ $$
Commented by M±th+et£s last updated on 26/Mar/20
$$\Leftrightarrow\sqrt{{x}}=\frac{{x}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 26/Mar/20
$${solution}\:{not}\:{correct}\:\left({complete}\right)\:{sir}! \\ $$$$\lfloor{A}\rfloor=\lfloor{B}\rfloor\:{doesn}'{t}\:{mean}\:{A}={B}\:! \\ $$$${example}:\:{A}=\mathrm{1}.\mathrm{2},\:{B}=\mathrm{1}.\mathrm{7} \\ $$$$\lfloor{A}\rfloor=\lfloor{B}\rfloor,\:{but}\:{A}\neq{B} \\ $$
Commented by mr W last updated on 26/Mar/20
$$\lfloor\sqrt{\mathrm{3}}\rfloor=\lfloor\frac{\mathrm{3}}{\mathrm{2}}\rfloor\:\Rightarrow{x}=\mathrm{3}\:{is}\:{also}\:{solution},{etc}. \\ $$
Commented by M±th+et£s last updated on 27/Mar/20
$${thank}\:{you}\:{sir}\:{but}\:{if}\:{we}\:{say} \\ $$$$\lfloor{A}\rfloor=\lfloor{B}\rfloor\:{means}\:{A}={B}\:{the}\:{solution}\:{will}\:{be} \\ $$$${wrong}\:{or}\:{not}\:{complete}\:{and}\:{why}\: \\ $$
Commented by mr W last updated on 27/Mar/20
$${if}\:\lfloor{A}\rfloor=\lfloor{B}\rfloor\:{it}\:{means} \\ $$$${A}={n}+{d}\:{with}\:\mathrm{0}\leqslant{d}<\mathrm{1} \\ $$$${B}={n}+\delta\:{with}\:\mathrm{0}\leqslant\delta<\mathrm{1} \\ $$$${if}\:{you}\:{say}\:{A}={B},\:{it}\:{is}\:{wrong}\:{in}\:{the} \\ $$$${sense}\:{of}\:{mathematics}. \\ $$$${if}\:{A}={B}\:\Rightarrow\lfloor{A}\rfloor=\lfloor{B}\rfloor \\ $$$${but}\:{if}\:\lfloor{A}\rfloor=\lfloor{B}\rfloor\nRightarrow{A}={B} \\ $$
Commented by M±th+et£s last updated on 27/Mar/20
$${thank}\:{you}\:{sir}\: \\ $$
Answered by Rio Michael last updated on 26/Mar/20
$$\:\lfloor\sqrt{{x}}\:\rfloor\:=\:\lfloor\frac{{x}}{\mathrm{2}}\rfloor \\ $$$$\:\Rightarrow\:\sqrt{{x}}\:=\:\frac{{x}}{\mathrm{2}} \\ $$$$\:\:\:{x}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:{x}^{\mathrm{2}} −\mathrm{4}{x}\:=\:\mathrm{0}\:\Leftrightarrow\:{x}\:=\:\mathrm{0}\:{or}\:{x}\:=\:\mathrm{4} \\ $$
Answered by mr W last updated on 27/Mar/20
$${x}\geqslant\mathrm{0} \\ $$$${let}\:\lfloor\:\sqrt{{x}}\:\rfloor=\lfloor\frac{{x}}{\mathrm{2}}\rfloor={n}\:\in{N} \\ $$$${n}=\mathrm{0}: \\ $$$$\Rightarrow\mathrm{0}\leqslant{x}<\mathrm{1}\:\Rightarrow{solution} \\ $$$$\Rightarrow\mathrm{0}\leqslant{x}<\mathrm{2} \\ $$$$ \\ $$$${n}=\mathrm{1}: \\ $$$$\Rightarrow\mathrm{1}\leqslant{x}<\mathrm{4} \\ $$$$\Rightarrow\mathrm{2}\leqslant{x}<\mathrm{4}\:\Rightarrow\:{solution} \\ $$$$ \\ $$$${n}=\mathrm{2}: \\ $$$$\Rightarrow\mathrm{4}\leqslant{x}<\mathrm{9} \\ $$$$\Rightarrow\mathrm{4}\leqslant{x}<\mathrm{6}\:\Rightarrow\:{solution} \\ $$$$ \\ $$$${n}=\mathrm{3}: \\ $$$$\Rightarrow\mathrm{9}\leqslant{x}<\mathrm{16} \\ $$$$\Rightarrow\mathrm{6}\leqslant{x}<\mathrm{8} \\ $$$$ \\ $$$$\Rightarrow{no}\:{solution}\:{upon}\:{n}\geqslant\mathrm{3} \\ $$$$ \\ $$$${solution}\:{is}: \\ $$$$\mathrm{0}\leqslant{x}<\mathrm{1}\:\vee\:\mathrm{2}\leqslant{x}<\mathrm{6} \\ $$
Commented by M±th+et£s last updated on 27/Mar/20
$${god}\:{bless}\:{you}\:{sir} \\ $$