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Question Number 175554 by Linton last updated on 02/Sep/22

x^(√x) =(√x^x )  find x

$${x}^{\sqrt{{x}}} =\sqrt{{x}^{{x}} } \\ $$$${find}\:{x} \\ $$

Answered by mr W last updated on 02/Sep/22

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

$${x}=\mathrm{0} \\ $$$${x}^{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} −\frac{{x}}{\mathrm{2}}} =\mathrm{1} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$$$\Rightarrow{x}^{\frac{\mathrm{1}}{\mathrm{2}}} −\frac{{x}}{\mathrm{2}}=\mathrm{0}\:\Rightarrow{x}=\mathrm{4} \\ $$

Commented by BaliramKumar last updated on 02/Sep/22

x = 0 undefiend

$${x}\:=\:\mathrm{0}\:\mathrm{undefiend} \\ $$

Commented by mr W last updated on 02/Sep/22

undefined for you, but not for all!

$${undefined}\:{for}\:{you},\:{but}\:{not}\:{for}\:{all}! \\ $$

Commented by mr W last updated on 02/Sep/22

Commented by BaliramKumar last updated on 02/Sep/22

you are right sir  3 solution valid

$${you}\:{are}\:{right}\:{sir} \\ $$$$\mathrm{3}\:{solution}\:{valid} \\ $$

Commented by BaliramKumar last updated on 02/Sep/22

  x^(√x)  = (√x^x )  x^x^(1/2)   = x^(x/2)   x^(1/2) log(x) = (x/2)log(x)  log(x){(√x) −(x/2)} = 0  log(x){2(√x)−x} = 0  log(x){2−(√x)}(√x) = 0  log_e (x) = 0        2−(√x) = 0        (√x) = 0  x = e^0                    (√x) = 2                  x = 0  x = 1                      x = 4                    x = 0

$$ \\ $$$${x}^{\sqrt{{x}}} \:=\:\sqrt{{x}^{{x}} } \\ $$$${x}^{{x}^{\frac{\mathrm{1}}{\mathrm{2}}} } \:=\:{x}^{\frac{{x}}{\mathrm{2}}} \\ $$$${x}^{\frac{\mathrm{1}}{\mathrm{2}}} {log}\left({x}\right)\:=\:\frac{{x}}{\mathrm{2}}{log}\left({x}\right) \\ $$$${log}\left({x}\right)\left\{\sqrt{{x}}\:−\frac{{x}}{\mathrm{2}}\right\}\:=\:\mathrm{0} \\ $$$${log}\left({x}\right)\left\{\mathrm{2}\sqrt{{x}}−{x}\right\}\:=\:\mathrm{0} \\ $$$${log}\left({x}\right)\left\{\mathrm{2}−\sqrt{{x}}\right\}\sqrt{{x}}\:=\:\mathrm{0} \\ $$$$\mathrm{log}_{\mathrm{e}} \left(\mathrm{x}\right)\:=\:\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{2}−\sqrt{\mathrm{x}}\:=\:\mathrm{0}\:\:\:\:\:\:\:\:\sqrt{\mathrm{x}}\:=\:\mathrm{0} \\ $$$$\mathrm{x}\:=\:\mathrm{e}^{\mathrm{0}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{\mathrm{x}}\:=\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\:=\:\mathrm{0} \\ $$$$\mathrm{x}\:=\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\:=\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\:=\:\mathrm{0} \\ $$

Commented by BaliramKumar last updated on 02/Sep/22

Answered by Rasheed.Sindhi last updated on 02/Sep/22

x^(√x) =(√x^x )  x^x^(1/2)  =x^(x/2)   x^(1/2) =x/2  x^(1/2−1) =1/2  (1/x^(1/2) )=(1/2)  x^(1/2) =2  x=2^2 =4

$${x}^{\sqrt{{x}}} =\sqrt{{x}^{{x}} } \\ $$$${x}^{{x}^{\mathrm{1}/\mathrm{2}} } ={x}^{{x}/\mathrm{2}} \\ $$$${x}^{\mathrm{1}/\mathrm{2}} ={x}/\mathrm{2} \\ $$$${x}^{\mathrm{1}/\mathrm{2}−\mathrm{1}} =\mathrm{1}/\mathrm{2} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{1}/\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{1}/\mathrm{2}} =\mathrm{2} \\ $$$${x}=\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$

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