Question-133180

Question Number 133180 by mathlove last updated on 19/Feb/21

Commented by mr W last updated on 19/Feb/21

no real solution! x^x ≥(1/( (e)^(1/e) ))≈0.692 (1/(256))<<0.692 ⇒x^x =(1/(256)) is not possible!

$${no}\:{real}\:{solution}! \\ $$$${x}^{{x}} \geqslant\frac{\mathrm{1}}{\:\sqrt[{{e}}]{{e}}}\approx\mathrm{0}.\mathrm{692} \\ $$$$\frac{\mathrm{1}}{\mathrm{256}}<<\mathrm{0}.\mathrm{692} \\ $$$$\Rightarrow{x}^{{x}} =\frac{\mathrm{1}}{\mathrm{256}}\:{is}\:{not}\:{possible}! \\ $$

Commented by mr W last updated on 20/Feb/21

generally the equation x^x =a has • no solution if a<(1/( (e)^(1/e) ))≈0.692 • one solution if a=(1/( (e)^(1/e) )) • two solutions if (1/( (e)^(1/e) ))<a<1 • one solution if a≥1 the solution(s) is (are): x=e^(W(ln a))

$${generally} \\ $$$${the}\:{equation}\:{x}^{{x}} ={a}\:{has} \\ $$$$\bullet\:{no}\:{solution}\:{if}\:{a}<\frac{\mathrm{1}}{\:\sqrt[{{e}}]{{e}}}\approx\mathrm{0}.\mathrm{692} \\ $$$$\bullet\:{one}\:{solution}\:{if}\:{a}=\frac{\mathrm{1}}{\:\sqrt[{{e}}]{{e}}} \\ $$$$\bullet\:{two}\:{solutions}\:{if}\:\frac{\mathrm{1}}{\:\sqrt[{{e}}]{{e}}}<{a}<\mathrm{1} \\ $$$$\bullet\:{one}\:{solution}\:{if}\:{a}\geqslant\mathrm{1} \\ $$$$ \\ $$$${the}\:{solution}\left({s}\right)\:{is}\:\left({are}\right): \\ $$$${x}={e}^{\mathbb{W}\left(\mathrm{ln}\:{a}\right)} \\ $$

Commented by Tawa11 last updated on 23/Jul/21

$$\mathrm{great} \\ $$

Answered by JDamian last updated on 19/Feb/21

$${x}\:=\:−\mathrm{4} \\ $$

Commented by liberty last updated on 20/Feb/21

Commented by mr W last updated on 20/Feb/21

4^(−4) =(1/(256)) but (−4)^((−4)) is not valid, so x=−4 is not correct.

$$\mathrm{4}^{−\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{256}} \\ $$$${but}\:\left(−\mathrm{4}\right)^{\left(−\mathrm{4}\right)} \:{is}\:{not}\:{valid},\:{so}\:{x}=−\mathrm{4}\:{is} \\ $$$${not}\:{correct}. \\ $$

Commented by liberty last updated on 20/Feb/21

in this graph clearly x^x = (1/(256)) no has solution

$$\mathrm{in}\:\mathrm{this}\:\mathrm{graph}\:\mathrm{clearly}\:\mathrm{x}^{\mathrm{x}} \:=\:\frac{\mathrm{1}}{\mathrm{256}} \\ $$$$\mathrm{no}\:\mathrm{has}\:\mathrm{solution} \\ $$

Commented by JDamian last updated on 20/Feb/21

The question is not about how the function x^x and its graph are. It is about which value of x powered to itself equals (1/(256)). No constraints about x values are given. You assumed x≥0 and so forgot to draw the point (−2,(1/4)) because (−2)^(−2) =(1/4) in the graph.

$${The}\:{question}\:{is}\:{not}\:{about}\:{how}\:{the} \\ $$$${function}\:{x}^{{x}} \:{and}\:{its}\:{graph}\:{are}. \\ $$$${It}\:{is}\:{about}\:{which}\:{value}\:{of}\:\:\:\boldsymbol{{x}}\:\:\:{powered} \\ $$$${to}\:{itself}\:{equals}\:\:\:\frac{\mathrm{1}}{\mathrm{256}}.\:{No}\:\:{constraints} \\ $$$${about}\:\:\boldsymbol{{x}}\:\:{values}\:{are}\:{given}. \\ $$$${You}\:{assumed}\:\:\:{x}\geqslant\mathrm{0}\:\:{and}\:{so}\:{forgot}\:{to}\:{draw} \\ $$$${the}\:{point}\:\left(−\mathrm{2},\frac{\mathrm{1}}{\mathrm{4}}\right)\:{because}\:\left(−\mathrm{2}\right)^{−\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${in}\:{the}\:{graph}. \\ $$

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