Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 113894 by Aina Samuel Temidayo last updated on 16/Sep/20

Commented by MJS_new last updated on 16/Sep/20

let x=2∧y=4 2^4 =4^2 =16 (x/y)^((x/y)) =(1/2)^((1/2)) =(1/( (√2)))=((√2)/2) x^((x/y−k)) =2^((1/2−k)) =((√2)/2^k )=((√2)/2) ⇒ k=1 let x=4∧y=2 4^2 +2^4 =16 (x/y)^((x/y)) =2^2 =4 x^((x/y−k)) =4^(2−k) =(4^2 /4^k )=4 ⇒ k=1

$$\mathrm{let}\:{x}=\mathrm{2}\wedge{y}=\mathrm{4} \\ $$$$\mathrm{2}^{\mathrm{4}} =\mathrm{4}^{\mathrm{2}} =\mathrm{16} \\ $$$$\left({x}/{y}\right)^{\left({x}/{y}\right)} =\left(\mathrm{1}/\mathrm{2}\right)^{\left(\mathrm{1}/\mathrm{2}\right)} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${x}^{\left({x}/{y}−{k}\right)} =\mathrm{2}^{\left(\mathrm{1}/\mathrm{2}−{k}\right)} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}^{{k}} }=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\:{k}=\mathrm{1} \\ $$$$\mathrm{let}\:{x}=\mathrm{4}\wedge{y}=\mathrm{2} \\ $$$$\mathrm{4}^{\mathrm{2}} +\mathrm{2}^{\mathrm{4}} =\mathrm{16} \\ $$$$\left({x}/{y}\right)^{\left({x}/{y}\right)} =\mathrm{2}^{\mathrm{2}} =\mathrm{4} \\ $$$${x}^{\left({x}/{y}−{k}\right)} =\mathrm{4}^{\mathrm{2}−{k}} =\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{4}^{{k}} }=\mathrm{4}\:\Rightarrow\:{k}=\mathrm{1} \\ $$

Answered by MJS_new last updated on 16/Sep/20

x^y =y^x ⇔ yln x =xln y ⇔ ((ln x)/x)=((ln y)/y) (x/y)^(x/y) =x^(x/y−k) (x^(x/y) /y^(x/y) )=(x^(x/y) /x^k ) y^(x/y) =x^k (x/y)ln y =kln x ((ln y)/y)=k((ln x)/y) ⇒ k=1

$${x}^{{y}} ={y}^{{x}} \:\Leftrightarrow\:{y}\mathrm{ln}\:{x}\:={x}\mathrm{ln}\:{y}\:\Leftrightarrow\:\frac{\mathrm{ln}\:{x}}{{x}}=\frac{\mathrm{ln}\:{y}}{{y}} \\ $$$$\left({x}/{y}\right)^{{x}/{y}} ={x}^{{x}/{y}−{k}} \\ $$$$\frac{{x}^{{x}/{y}} }{{y}^{{x}/{y}} }=\frac{{x}^{{x}/{y}} }{{x}^{{k}} } \\ $$$${y}^{{x}/{y}} ={x}^{{k}} \\ $$$$\frac{{x}}{{y}}\mathrm{ln}\:{y}\:={k}\mathrm{ln}\:{x} \\ $$$$\frac{\mathrm{ln}\:{y}}{{y}}={k}\frac{\mathrm{ln}\:{x}}{{y}}\:\Rightarrow\:{k}=\mathrm{1} \\ $$

Commented by Aina Samuel Temidayo last updated on 16/Sep/20

Sorry, I don′t understand your last two steps.

$$\mathrm{Sorry},\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{your}\:\mathrm{last} \\ $$$$\mathrm{two}\:\mathrm{steps}. \\ $$

Commented by MJS_new last updated on 16/Sep/20

ln (a^b ) =bln a ln (y^(x/y) ) =(x/y)ln y ln x^k =kln x (x/y)ln y =kln x divide both sides by x (1/y)ln y =((kln x)/c) same as ((ln y)/y)=k((ln x)/x) and in the first line I showed ((ln y)/y)=((ln x)/x) ⇒ k=1

$$\mathrm{ln}\:\left({a}^{{b}} \right)\:={b}\mathrm{ln}\:{a} \\ $$$$\mathrm{ln}\:\left({y}^{{x}/{y}} \right)\:=\frac{{x}}{{y}}\mathrm{ln}\:{y} \\ $$$$\mathrm{ln}\:{x}^{{k}} \:={k}\mathrm{ln}\:{x} \\ $$$$ \\ $$$$\frac{{x}}{{y}}\mathrm{ln}\:{y}\:={k}\mathrm{ln}\:{x}\:\mathrm{divide}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{by}\:{x} \\ $$$$\frac{\mathrm{1}}{{y}}\mathrm{ln}\:{y}\:=\frac{{k}\mathrm{ln}\:{x}}{{c}}\:\mathrm{same}\:\mathrm{as}\:\frac{\mathrm{ln}\:{y}}{{y}}={k}\frac{\mathrm{ln}\:{x}}{{x}} \\ $$$$\mathrm{and}\:\mathrm{in}\:\mathrm{the}\:\mathrm{first}\:\mathrm{line}\:\mathrm{I}\:\mathrm{showed}\:\frac{\mathrm{ln}\:{y}}{{y}}=\frac{\mathrm{ln}\:{x}}{{x}} \\ $$$$\Rightarrow\:{k}=\mathrm{1} \\ $$

Answered by john santu last updated on 16/Sep/20

(1)x^y =y^x ⇒x=y^(x/y) (2)((x/y))^(x/y) =(x)^((x/y)−k) ⇒ ((y^(y/x) /y))^(x/y) =(y^(x/y) )^((x/y)−k) (y^((y/x)−1) )^(x/y) =(y)^((x/y)((x/y)−k)) ⇒ (x/y)((y/x)−1)=(x/y)((x/y)−k) ⇒(y/x)−1=(x/y)−k ⇒k = (x/y)−(y/x)+1=((x^2 −y^2 +xy)/(xy)) ⇒k = (((x+y)^2 −2y^2 −xy)/(xy)) ⇒k=(((x+y+y(√2))(x+y−y(√2))−xy)/(xy))

$$\left(\mathrm{1}\right){x}^{{y}} ={y}^{{x}} \Rightarrow{x}={y}^{\frac{{x}}{{y}}} \\ $$$$\left(\mathrm{2}\right)\left(\frac{{x}}{{y}}\right)^{\frac{{x}}{{y}}} =\left({x}\right)^{\frac{{x}}{{y}}−{k}} \\ $$$$\Rightarrow\:\left(\frac{{y}^{\frac{{y}}{{x}}} }{{y}}\right)^{\frac{{x}}{{y}}} =\left({y}^{\frac{{x}}{{y}}} \right)^{\frac{{x}}{{y}}−{k}} \\ $$$$\left({y}^{\frac{{y}}{{x}}−\mathrm{1}} \right)^{\frac{{x}}{{y}}} =\left({y}\right)^{\frac{{x}}{{y}}\left(\frac{{x}}{{y}}−{k}\right)} \\ $$$$\Rightarrow\:\frac{{x}}{{y}}\left(\frac{{y}}{{x}}−\mathrm{1}\right)=\frac{{x}}{{y}}\left(\frac{{x}}{{y}}−{k}\right) \\ $$$$\Rightarrow\frac{{y}}{{x}}−\mathrm{1}=\frac{{x}}{{y}}−{k}\: \\ $$$$\Rightarrow{k}\:=\:\frac{{x}}{{y}}−\frac{{y}}{{x}}+\mathrm{1}=\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +{xy}}{{xy}} \\ $$$$\Rightarrow{k}\:=\:\frac{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} −{xy}}{{xy}} \\ $$$$\Rightarrow{k}=\frac{\left({x}+{y}+{y}\sqrt{\mathrm{2}}\right)\left({x}+{y}−{y}\sqrt{\mathrm{2}}\right)−{xy}}{{xy}} \\ $$

Commented by MJS_new last updated on 16/Sep/20

x=y solves x^y =y^x in this case your k=(x/y)−(y/x)+1=1

$${x}={y}\:\mathrm{solves}\:{x}^{{y}} ={y}^{{x}} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{your}\:{k}=\frac{{x}}{{y}}−\frac{{y}}{{x}}+\mathrm{1}=\mathrm{1} \\ $$

Commented by MJS_new last updated on 16/Sep/20

but with x=2∧y=4 ⇒ k=−(1/2) and with x=4∧y=2 ⇒ k=(5/2); both are wrong

$$\mathrm{but}\:\mathrm{with}\:{x}=\mathrm{2}\wedge{y}=\mathrm{4}\:\Rightarrow\:{k}=−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{with} \\ $$$${x}=\mathrm{4}\wedge{y}=\mathrm{2}\:\Rightarrow\:{k}=\frac{\mathrm{5}}{\mathrm{2}};\:\mathrm{both}\:\mathrm{are}\:\mathrm{wrong} \\ $$

Commented by bobhans last updated on 16/Sep/20

it means x must be equal to y?

$${it}\:{means}\:{x}\:{must}\:{be}\:{equal}\:{to}\:{y}? \\ $$

Commented by MJS_new last updated on 16/Sep/20

no, see my answer above

$$\mathrm{no},\:\mathrm{see}\:\mathrm{my}\:{answer}\:\mathrm{above} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com