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Question Number 201200 by Calculusboy last updated on 01/Dec/23

Commented by Rasheed.Sindhi last updated on 02/Dec/23

( (x−1)^(x−1) )^(1/3) =(x−1)^((x−1)^(1/3) )

$$\left(\:\left({x}−\mathrm{1}\right)^{{x}−\mathrm{1}} \:\right)^{\mathrm{1}/\mathrm{3}} =\left({x}−\mathrm{1}\right)^{\left({x}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$

Commented by Rasheed.Sindhi last updated on 02/Dec/23

Symmetry in above question!

$${Symmetry}\:{in}\:{above}\:{question}! \\ $$

Answered by cortano12 last updated on 02/Dec/23

(x−1)^((x−1)/3) = (x−1)^((x−1))^(1/3) ; x≠1 (1) x−1=−1⇒x=0 (2) x−1=1⇒x=2 (3) ((x−1)/3) =((x−1))^(1/3) ⇒(x−1)=3(x−1)^(1/3) (x−1)^(1/3) ((x−1)^(2/3) −3)=0 (x−1)^(2/3) = 3⇒(x−1)^2 =27 x= 1±3(√3)

$$\:\left(\mathrm{x}−\mathrm{1}\right)^{\frac{\mathrm{x}−\mathrm{1}}{\mathrm{3}}} \:=\:\left(\mathrm{x}−\mathrm{1}\right)^{\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{1}}} \:;\:\mathrm{x}\neq\mathrm{1} \\ $$$$\:\left(\mathrm{1}\right)\:\mathrm{x}−\mathrm{1}=−\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{x}−\mathrm{1}=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{2} \\ $$$$\:\left(\mathrm{3}\right)\:\frac{\mathrm{x}−\mathrm{1}}{\mathrm{3}}\:=\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{1}}\Rightarrow\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{3}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \:\left(\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}/\mathrm{3}} −\mathrm{3}\right)=\mathrm{0} \\ $$$$\:\:\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}/\mathrm{3}} =\:\mathrm{3}\Rightarrow\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{27} \\ $$$$\:\:\:\mathrm{x}=\:\mathrm{1}\pm\mathrm{3}\sqrt{\mathrm{3}}\: \\ $$

Commented by Calculusboy last updated on 02/Dec/23

thanks sir

$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

Answered by esmaeil last updated on 02/Dec/23

((x−1)/3)ln(x−1)=((x−1))^(1/3) ln(x−1)→^(x−1=u) u=3(u)^(1/3) → { ((x=1)),((x=1±3(√3))) :}

$$\frac{{x}−\mathrm{1}}{\mathrm{3}}{ln}\left({x}−\mathrm{1}\right)=\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}{ln}\left({x}−\mathrm{1}\right)\overset{{x}−\mathrm{1}={u}} {\rightarrow} \\ $$$${u}=\mathrm{3}\sqrt[{\mathrm{3}}]{{u}}\rightarrow\begin{cases}{{x}=\mathrm{1}}\\{\mathrm{x}=\mathrm{1}\pm\mathrm{3}\sqrt{\mathrm{3}}}\end{cases} \\ $$

Commented by Calculusboy last updated on 02/Dec/23

thanks

$$\boldsymbol{{thanks}} \\ $$

Answered by Rasheed.Sindhi last updated on 02/Dec/23

(x−1)^((x−1)/3) =(x−1)^((x−1)^(1/3) ) (x−1)^(((x−1)/3)−(x−1)^(1/3) ) =1=(x−1)^0 { ((x−1=1⇒x=2✓)),((x−1=−1⇒x=0✓)),((((x−1)/3)−(x−1)^(1/3) =0)) :} ((x−1−3(x−1)^(1/3) )/3)=0 x−1=3(x−1)^(1/3) (x−1)^3 =27(x−1) (x−1)^3 −27(x−1)=0 (x−1)( (x−1)^2 −27)=0 x≠1⇒ x−1=±3(√3) ⇒x=1±3(√3) ✓

$$\left({x}−\mathrm{1}\right)^{\frac{{x}−\mathrm{1}}{\mathrm{3}}} =\left({x}−\mathrm{1}\right)^{\left({x}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$$\left({x}−\mathrm{1}\right)^{\frac{{x}−\mathrm{1}}{\mathrm{3}}−\left({x}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} } =\mathrm{1}=\left({x}−\mathrm{1}\right)^{\mathrm{0}} \\ $$$$\begin{cases}{{x}−\mathrm{1}=\mathrm{1}\Rightarrow{x}=\mathrm{2}\checkmark}\\{{x}−\mathrm{1}=−\mathrm{1}\Rightarrow{x}=\mathrm{0}\checkmark}\\{\frac{{x}−\mathrm{1}}{\mathrm{3}}−\left({x}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{0}}\end{cases} \\ $$$$\frac{{x}−\mathrm{1}−\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} }{\mathrm{3}}=\mathrm{0} \\ $$$${x}−\mathrm{1}=\mathrm{3}\left({x}−\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{3}} =\mathrm{27}\left({x}−\mathrm{1}\right) \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{3}} −\mathrm{27}\left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left(\:\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{27}\right)=\mathrm{0} \\ $$$${x}\neq\mathrm{1}\Rightarrow\:\:{x}−\mathrm{1}=\pm\mathrm{3}\sqrt{\mathrm{3}}\:\Rightarrow{x}=\mathrm{1}\pm\mathrm{3}\sqrt{\mathrm{3}}\:\:\checkmark\:\:\:\:\:\: \\ $$

Commented by Calculusboy last updated on 02/Dec/23

nice solution

$$\boldsymbol{{nice}}\:\boldsymbol{{solution}} \\ $$

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