Question-194869

Question Number 194869 by sonukgindia last updated on 17/Jul/23

Answered by Frix last updated on 17/Jul/23

−x^2^(−x^4 ) =−x^4^(−x^2 ) x^((1/2)^x^4 ) =x^((1/4)^x^2 ) (1/2)^x^4 ln x =(1/4)^x^2 ln x ⇒ x=1 (1/2)^x^4 =(1/4)^x^2 x^4 ln (1/2) =x^2 ln (1/4) −x^4 ln 2 =−2x^2 ln 2 x^4 =2x^2 ⇒ x=0 x^2 =2 ⇒ x=±(√2)

$$−{x}^{\mathrm{2}^{−{x}^{\mathrm{4}} } } =−{x}^{\mathrm{4}^{−{x}^{\mathrm{2}} } } \\ $$$${x}^{\left(\mathrm{1}/\mathrm{2}\right)^{{x}^{\mathrm{4}} } } ={x}^{\left(\mathrm{1}/\mathrm{4}\right)^{{x}^{\mathrm{2}} } } \\ $$$$\left(\mathrm{1}/\mathrm{2}\right)^{{x}^{\mathrm{4}} } \mathrm{ln}\:{x}\:=\left(\mathrm{1}/\mathrm{4}\right)^{{x}^{\mathrm{2}} } \mathrm{ln}\:{x}\:\Rightarrow\:{x}=\mathrm{1} \\ $$$$\left(\mathrm{1}/\mathrm{2}\right)^{{x}^{\mathrm{4}} } =\left(\mathrm{1}/\mathrm{4}\right)^{{x}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{4}} \mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}}\:={x}^{\mathrm{2}} \mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$−{x}^{\mathrm{4}} \mathrm{ln}\:\mathrm{2}\:=−\mathrm{2}{x}^{\mathrm{2}} \mathrm{ln}\:\mathrm{2} \\ $$$${x}^{\mathrm{4}} =\mathrm{2}{x}^{\mathrm{2}} \:\Rightarrow\:{x}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\mathrm{2}\:\Rightarrow\:{x}=\pm\sqrt{\mathrm{2}} \\ $$

Commented by Frix last updated on 18/Jul/23

−x^2^(−x^4 ) =−(x↑(2↑(−(x↑4)))) −x^4^(−x^2 ) =−(x↑(4↑(−(x↑2)))) Both are identical for x=−(√2) −(−(√2))^4 =−4 2^(−4) =(1/(16)) −(−(√2))^(1/(16)) =−2^(1/(32)) i −(−(√2))^2 =−2 4^(−2) =(1/(16)) −(−(√2))^(1/(16)) =−2^(1/(32)) i It says “x∈R” and −(√2)∈R It′s like Solve for x∈R: (1+4i)^x =(3−2i)^x ⇒ x=0∈R

$$−{x}^{\mathrm{2}^{−{x}^{\mathrm{4}} } } =−\left({x}\uparrow\left(\mathrm{2}\uparrow\left(−\left({x}\uparrow\mathrm{4}\right)\right)\right)\right) \\ $$$$−{x}^{\mathrm{4}^{−{x}^{\mathrm{2}} } } =−\left({x}\uparrow\left(\mathrm{4}\uparrow\left(−\left({x}\uparrow\mathrm{2}\right)\right)\right)\right) \\ $$$$\mathrm{Both}\:\mathrm{are}\:\mathrm{identical}\:\mathrm{for}\:{x}=−\sqrt{\mathrm{2}} \\ $$$$−\left(−\sqrt{\mathrm{2}}\right)^{\mathrm{4}} =−\mathrm{4}\:\:\:\:\:\mathrm{2}^{−\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{16}}\:\:\:\:\:−\left(−\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{16}}} =−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{32}}} \mathrm{i} \\ $$$$−\left(−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =−\mathrm{2}\:\:\:\:\:\mathrm{4}^{−\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{16}}\:\:\:\:\:−\left(−\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{16}}} =−\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{32}}} \mathrm{i} \\ $$$$\mathrm{It}\:\mathrm{says}\:“{x}\in\mathbb{R}''\:\mathrm{and}\:−\sqrt{\mathrm{2}}\in\mathbb{R} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{like} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:{x}\in\mathbb{R}:\:\left(\mathrm{1}+\mathrm{4i}\right)^{{x}} =\left(\mathrm{3}−\mathrm{2i}\right)^{{x}} \:\Rightarrow\:{x}=\mathrm{0}\in\mathbb{R} \\ $$

Commented by aba last updated on 18/Jul/23

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