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Question Number 172005 by Mikenice last updated on 23/Jun/22
$${find}\:{x}:\:\left({log}_{\mathrm{10}} {x}\right)^{\mathrm{2}} −{log}_{\mathrm{10}} {x}=\mathrm{0} \\ $$
Answered by puissant last updated on 23/Jun/22
$${X}={log}_{\mathrm{10}} {x}\:\:\:\Rightarrow\:\:{X}^{\mathrm{2}} −{X}=\mathrm{0} \\ $$$$\Rightarrow\:{X}\left({X}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{X}=\mathrm{0}\:\:\:{or}\:\:{X}=\mathrm{1} \\ $$$${log}_{\mathrm{10}} {x}\:=\:\frac{{lnx}}{{ln}\mathrm{10}}=\mathrm{0}\:\Rightarrow\:{lnx}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{1} \\ $$$${log}_{\mathrm{10}} {x}=\frac{{lnx}}{{ln}\mathrm{10}}=\mathrm{1}\:\Rightarrow\:{lnx}={ln}\mathrm{10}\:\Rightarrow\:{x}=\mathrm{10}. \\ $$
Answered by Rasheed.Sindhi last updated on 23/Jun/22
$${find}\:{x}:\:\left({log}_{\mathrm{10}} {x}\right)^{\mathrm{2}} −{log}_{\mathrm{10}} {x}=\mathrm{0} \\ $$$${log}_{\mathrm{10}} {x}\:\left({log}_{\mathrm{10}} {x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${log}_{\mathrm{10}} {x}=\mathrm{0}\:\mid\:{log}_{\mathrm{10}} {x}=\mathrm{1} \\ $$$$\:\:\:{x}=\mathrm{10}^{\mathrm{0}} \:\:\mid\:\:{x}=\mathrm{10}^{\mathrm{1}} \\ $$$${x}=\mathrm{1},\:\mathrm{10} \\ $$
Commented by Mikenice last updated on 23/Jun/22
$${yhanks}\:{sir} \\ $$