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Question Number 100150 by student work last updated on 25/Jun/20

x^(ln) −e^6 ∙x=0 x=? help me

$$\mathrm{x}^{\mathrm{ln}} −\mathrm{e}^{\mathrm{6}} \centerdot\mathrm{x}=\mathrm{0}\:\:\:\:\:\:\mathrm{x}=? \\ $$$$\mathrm{help}\:\mathrm{me} \\ $$

Commented by student work last updated on 25/Jun/20

x^(lnx) −e^6 ∙x=0

$$\mathrm{x}^{\mathrm{lnx}} −\mathrm{e}^{\mathrm{6}} \centerdot\mathrm{x}=\mathrm{0} \\ $$

Commented by bobhans last updated on 25/Jun/20

your question not clear

$$\mathrm{your}\:\mathrm{question}\:\mathrm{not}\:\mathrm{clear} \\ $$

Commented by bemath last updated on 25/Jun/20

ln(x^(ln x) ) = ln(e^6 .x) (ln(x))^2 = 6+ln(x) p^2 −p−6 =0 , p = ln(x) (p−3)(p+2)=0 p=3 ⇒ln(x)=3 , x = e^3 p=−2⇒ln(x)=−2, x = e^(−2)

$$\mathrm{ln}\left(\mathrm{x}^{\mathrm{ln}\:\mathrm{x}} \right)\:=\:\mathrm{ln}\left(\mathrm{e}^{\mathrm{6}} .\mathrm{x}\right) \\ $$$$\left(\mathrm{ln}\left(\mathrm{x}\right)\right)^{\mathrm{2}} =\:\mathrm{6}+\mathrm{ln}\left(\mathrm{x}\right) \\ $$$$\mathrm{p}^{\mathrm{2}} −\mathrm{p}−\mathrm{6}\:=\mathrm{0}\:,\:\mathrm{p}\:=\:\mathrm{ln}\left(\mathrm{x}\right) \\ $$$$\left(\mathrm{p}−\mathrm{3}\right)\left(\mathrm{p}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{p}=\mathrm{3}\:\Rightarrow\mathrm{ln}\left(\mathrm{x}\right)=\mathrm{3}\:,\:\mathrm{x}\:=\:\mathrm{e}^{\mathrm{3}} \\ $$$$\mathrm{p}=−\mathrm{2}\Rightarrow\mathrm{ln}\left(\mathrm{x}\right)=−\mathrm{2},\:\mathrm{x}\:=\:\mathrm{e}^{−\mathrm{2}} \\ $$

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