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Question Number 66890 by hmamarques1994@gmail.com last updated on 20/Aug/19
x^x + x^(2x) = 20 x^x = ?
$$\: \\ $$$$\:\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \:+\:\boldsymbol{\mathrm{x}}^{\mathrm{2}\boldsymbol{\mathrm{x}}} \:=\:\mathrm{20} \\ $$$$\: \\ $$$$\:\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \:=\:? \\ $$$$\: \\ $$
Answered by Kunal12588 last updated on 21/Aug/19
let x^x =t t+t^2 =20 ⇒t^2 +t−20=0 ⇒t=((−1±(√(1+80)))/2)=((−1±9)/2) t_1 =4,t_2 =−5(invalid) ⇒x^x =4
$${let}\:{x}^{{x}} ={t} \\ $$$${t}+{t}^{\mathrm{2}} =\mathrm{20} \\ $$$$\Rightarrow{t}^{\mathrm{2}} +{t}−\mathrm{20}=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{80}}}{\mathrm{2}}=\frac{−\mathrm{1}\pm\mathrm{9}}{\mathrm{2}} \\ $$$${t}_{\mathrm{1}} =\mathrm{4},{t}_{\mathrm{2}} =−\mathrm{5}\left({invalid}\right) \\ $$$$\Rightarrow{x}^{{x}} =\mathrm{4} \\ $$
Answered by John Kaloki Musau last updated on 20/Aug/19
x^x +x^x ×x^x =20 let x^x be y y+y^2 =20 y^2 +y−20=0 y^2 +5y−4y−20=0 y(y+5)−4(y+5)=0 (y−4)(y+5)=0 y=4 or y=−5 x^x =4 or −5
$${x}^{{x}} +{x}^{{x}} ×{x}^{{x}} =\mathrm{20} \\ $$$${let}\:{x}^{{x}} \:{be}\:{y} \\ $$$${y}+{y}^{\mathrm{2}} =\mathrm{20} \\ $$$${y}^{\mathrm{2}} +{y}−\mathrm{20}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} +\mathrm{5}{y}−\mathrm{4}{y}−\mathrm{20}=\mathrm{0} \\ $$$${y}\left({y}+\mathrm{5}\right)−\mathrm{4}\left({y}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\left({y}−\mathrm{4}\right)\left({y}+\mathrm{5}\right)=\mathrm{0} \\ $$$${y}=\mathrm{4}\:{or}\:{y}=−\mathrm{5} \\ $$$${x}^{{x}} =\mathrm{4}\:{or}\:−\mathrm{5} \\ $$
Answered by mr W last updated on 20/Aug/19
let t=x^x >0 t+t^2 =20 (t−4)(t+5)=0 ⇒t=4, −5 since t>0, only t=4 is ok. ⇒x^x =4 ⇒x=2
$${let}\:{t}={x}^{{x}} >\mathrm{0} \\ $$$${t}+{t}^{\mathrm{2}} =\mathrm{20} \\ $$$$\left({t}−\mathrm{4}\right)\left({t}+\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{4},\:−\mathrm{5} \\ $$$${since}\:{t}>\mathrm{0},\:{only}\:{t}=\mathrm{4}\:{is}\:{ok}. \\ $$$$\Rightarrow{x}^{{x}} =\mathrm{4} \\ $$$$\Rightarrow{x}=\mathrm{2} \\ $$
Commented by John Kaloki Musau last updated on 21/Aug/19
t is a number and can take both positive and negative forms.I′m I correct?
$${t}\:{is}\:{a}\:{number}\:{and}\:{can}\:{take}\:{both}\:{positive} \\ $$$${and}\:{negative}\:{forms}.{I}'{m}\:{I}\:{correct}? \\ $$
Commented by mr W last updated on 21/Aug/19
but t stands for x^x which is always positive, therefore t is positive.
$${but}\:{t}\:{stands}\:{for}\:{x}^{{x}} \:{which}\:{is}\:{always} \\ $$$${positive},\:{therefore}\:{t}\:{is}\:{positive}. \\ $$

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