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

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

Answered by MJS_new last updated on 16/Sep/20

$$x^y=y^x\iff y\ln x=x\ln y\iff \frac{\ln x}{x}=\frac{\ln y}{y}$$$${\left(x/y\right)}^{x/y}=x^{x/y-k}$$$$\frac{x^{x/y}}{y^{x/y}}=\frac{x^{x/y}}{x^k}$$$$y^{x/y}=x^k$$$$\frac{x}{y}\ln y=k\ln x$$$$\frac{\ln y}{y}=k\frac{\ln x}{y}\Rightarrow k=1$$

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

$$\mathrm{Sorry},\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{your}\mathrm{last}$$$$\mathrm{two}\mathrm{steps}.$$

Commented by MJS_new last updated on 16/Sep/20

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

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

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

Commented by MJS_new last updated on 16/Sep/20

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

Commented by MJS_new last updated on 16/Sep/20

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

Commented by bobhans last updated on 16/Sep/20

$$itmeansxmustbeequaltoy?$$

Commented by MJS_new last updated on 16/Sep/20

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com