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Question Number 32897 by artibunja last updated on 06/Apr/18

e^π > π^e or e^π <π^e ?

$$ \\ $$ $${e}^{\pi} \:>\:\pi^{{e}} \:\:{or}\:\:\:{e}^{\pi} <\pi^{{e}} \:? \\ $$ $$ \\ $$

Answered by MJS last updated on 06/Apr/18

e^x =x^e ; x>1 ⇒ x=e ⇒ ∀x>e: e^x >x^e π>e ⇒ e^π >π^e

$${e}^{{x}} ={x}^{{e}} ;\:{x}>\mathrm{1}\:\Rightarrow\:{x}={e}\:\Rightarrow\:\forall{x}>{e}:\:{e}^{{x}} >{x}^{{e}} \\ $$ $$\pi>{e}\:\Rightarrow\:{e}^{\pi} >\pi^{{e}} \\ $$

Commented byMJS last updated on 06/Apr/18

interestingly for a^x =x^a with a>1 and x>1 two solutions seem to exist (i.e. 2^x =x^2 ⇒ x=2∨x=4; 3^x =x^3 ⇒ x=3∨x≈2.47805) but for a=e there′s only one: x=e can anybody further explain this?

$$\mathrm{interestingly}\:\mathrm{for}\:{a}^{{x}} ={x}^{{a}} \:\mathrm{with}\:{a}>\mathrm{1} \\ $$ $$\mathrm{and}\:{x}>\mathrm{1}\:\mathrm{two}\:\mathrm{solutions}\:\mathrm{seem}\:\mathrm{to} \\ $$ $$\mathrm{exist}\:\left(\mathrm{i}.\mathrm{e}.\:\mathrm{2}^{{x}} ={x}^{\mathrm{2}} \:\Rightarrow\:{x}=\mathrm{2}\vee{x}=\mathrm{4};\right. \\ $$ $$\left.\mathrm{3}^{{x}} ={x}^{\mathrm{3}} \:\Rightarrow\:{x}=\mathrm{3}\vee{x}\approx\mathrm{2}.\mathrm{47805}\right)\:\mathrm{but} \\ $$ $$\mathrm{for}\:{a}={e}\:\mathrm{there}'\mathrm{s}\:\mathrm{only}\:\mathrm{one}:\:{x}={e} \\ $$ $$ \\ $$ $$\mathrm{can}\:\mathrm{anybody}\:\mathrm{further}\:\mathrm{explain}\:\mathrm{this}? \\ $$

Commented bymrW2 last updated on 06/Apr/18

solution for a^x =x^a (a>0): a^(x/a) =x e^((x/a) ln a) =x (−(x/a) ln a) e^(−(x/a) ln a) =−((ln a)/a) ⇒−(x/a) ln a=W(−((ln a)/a)) ⇒x=−(a/(ln a)) W(−((ln a)/a))=−((W(−ln a^(1/a) ))/(ln a^(1/a) )) if a=e: one solution if a≠e: two solutions

$${solution}\:{for}\:{a}^{{x}} ={x}^{{a}} \:\left({a}>\mathrm{0}\right): \\ $$ $${a}^{\frac{{x}}{{a}}} ={x} \\ $$ $${e}^{\frac{{x}}{{a}}\:\mathrm{ln}\:{a}} ={x} \\ $$ $$\left(−\frac{{x}}{{a}}\:\mathrm{ln}\:{a}\right)\:{e}^{−\frac{{x}}{{a}}\:\mathrm{ln}\:{a}} =−\frac{\mathrm{ln}\:{a}}{{a}} \\ $$ $$\Rightarrow−\frac{{x}}{{a}}\:\mathrm{ln}\:{a}={W}\left(−\frac{\mathrm{ln}\:{a}}{{a}}\right) \\ $$ $$\Rightarrow{x}=−\frac{{a}}{\mathrm{ln}\:{a}}\:{W}\left(−\frac{\mathrm{ln}\:{a}}{{a}}\right)=−\frac{{W}\left(−\mathrm{ln}\:{a}^{\frac{\mathrm{1}}{{a}}} \right)}{\mathrm{ln}\:{a}^{\frac{\mathrm{1}}{{a}}} } \\ $$ $${if}\:{a}={e}:\:{one}\:{solution} \\ $$ $${if}\:{a}\neq{e}:\:{two}\:{solutions} \\ $$

Answered by Tinkutara last updated on 06/Apr/18

Consider f(x)=y=x^(1/x) (dy/dx)=(y/x^2 )(1−ln x) If x<e, f(x) is increasing. If x>e, f(x) is decreasing. ∴ e^(1/e) >π^(1/π) e^π >π^e

$${Consider}\:{f}\left({x}\right)={y}={x}^{\frac{\mathrm{1}}{{x}}} \\ $$ $$\frac{{dy}}{{dx}}=\frac{{y}}{{x}^{\mathrm{2}} }\left(\mathrm{1}−\mathrm{ln}\:{x}\right) \\ $$ $${If}\:{x}<{e},\:{f}\left({x}\right)\:{is}\:{increasing}. \\ $$ $${If}\:{x}>{e},\:{f}\left({x}\right)\:{is}\:{decreasing}. \\ $$ $$\therefore\:{e}^{\frac{\mathrm{1}}{{e}}} >\pi^{\frac{\mathrm{1}}{\pi}} \\ $$ $${e}^{\pi} >\pi^{{e}} \\ $$

Answered by artibunja last updated on 06/Apr/18

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