Question Number 87637 by john santu last updated on 05/Apr/20
$$\sqrt[{\mathrm{4}\:\:}]{\mid\mathrm{x}−\mathrm{3}\mid^{\mathrm{x}+\mathrm{1}} }\:=\:\sqrt[{\mathrm{3}\:\:}]{\mid\mathrm{x}−\mathrm{3}\mid^{\mathrm{x}−\mathrm{2}} } \\ $$
Answered by TANMAY PANACEA. last updated on 05/Apr/20
$$\mid{x}−\mathrm{3}\mid^{\frac{{x}+\mathrm{1}}{\mathrm{4}}} =\mid{x}−\mathrm{3}\mid^{\frac{{x}−\mathrm{2}}{\mathrm{3}}} \\ $$$$\frac{{x}+\mathrm{1}}{\mathrm{4}}=\frac{{x}−\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{4}{x}−\mathrm{8}=\mathrm{3}{x}+\mathrm{3} \\ $$$${x}=\mathrm{11} \\ $$
Commented by john santu last updated on 05/Apr/20
$$\mathrm{has}\:\mathrm{three}\:\mathrm{solution}\:\mathrm{sir} \\ $$$$\mathrm{x}\:=\:\mathrm{2};\:\mathrm{4}\:;\:\mathrm{11} \\ $$
Commented by TANMAY PANACEA. last updated on 05/Apr/20
$${ok} \\ $$
Commented by Rasheed.Sindhi last updated on 05/Apr/20
$${What}\:{about}\:{x}=\mathrm{3}?\:{Isn}'{t}\:{it}\:{root}? \\ $$
Commented by john santu last updated on 05/Apr/20
$$\mathrm{no} \\ $$
Commented by Rasheed.Sindhi last updated on 05/Apr/20
$${But}\:{x}=\mathrm{3}\:{satisfies}\:{the}\:{given}\:{eqn}. \\ $$$$\:^{\mathrm{4}} \sqrt{\mid{x}−\mathrm{3}\mid^{{x}+\mathrm{1}} }=\:^{\mathrm{3}} \sqrt{\mid{x}−\mathrm{3}\mid^{{x}−\mathrm{2}} } \\ $$$$\:^{\mathrm{4}} \sqrt{\mid\mathrm{3}−\mathrm{3}\mid^{\mathrm{3}+\mathrm{1}} }=\:^{\mathrm{3}} \sqrt{\mid\mathrm{3}−\mathrm{3}\mid^{\mathrm{3}−\mathrm{2}} } \\ $$$$\:^{\mathrm{4}} \sqrt{\mathrm{0}^{\mathrm{4}} }=\:^{\mathrm{3}} \sqrt{\mathrm{0}^{\mathrm{1}} } \\ $$$$\:^{\mathrm{4}} \sqrt{\mathrm{0}}=\:^{\mathrm{3}} \sqrt{\mathrm{0}} \\ $$$$\:\:\:\:\mathrm{0}=\mathrm{0} \\ $$