Menu Close

if-a-a-b-b-a-b-is-it-true-if-it-is-true-then-prove-it-

Tinku Tara 0 Comments
Pin




Question Number 205849 by Davidtim last updated on 31/Mar/24
if a^a =b^b ; a=b is it true? if it is true then prove it.
$${if}\:{a}^{{a}} ={b}^{{b}} \:\:;\:{a}={b}\:\:{is}\:{it}\:{true}? \\ $$$${if}\:{it}\:{is}\:{true}\:{then}\:{prove}\:{it}. \\ $$
Commented by mr W last updated on 31/Mar/24
not true for 0<a, b<1. example: a, b=((ln 0.8)/(W(ln 0.8)))≈ { ((0.094649710865)),((0.739533650011)) :} 0.094649710865^(0.094649710865) =0.739533650011^(0.739533650011) =0.8
$${not}\:{true}\:{for}\:\mathrm{0}<{a},\:{b}<\mathrm{1}. \\ $$$${example}: \\ $$$${a},\:{b}=\frac{\mathrm{ln}\:\mathrm{0}.\mathrm{8}}{{W}\left(\mathrm{ln}\:\mathrm{0}.\mathrm{8}\right)}\approx\begin{cases}{\mathrm{0}.\mathrm{094649710865}}\\{\mathrm{0}.\mathrm{739533650011}}\end{cases} \\ $$$$\mathrm{0}.\mathrm{094649710865}^{\mathrm{0}.\mathrm{094649710865}} \\ $$$$=\mathrm{0}.\mathrm{739533650011}^{\mathrm{0}.\mathrm{739533650011}} \\ $$$$=\mathrm{0}.\mathrm{8} \\ $$
Commented by Davidtim last updated on 01/Apr/24
please check the above exp.
$${please}\:{check}\:{the}\:{above}\:{exp}. \\ $$
Commented by Davidtim last updated on 01/Apr/24
Commented by mr W last updated on 01/Apr/24
if ((x/3))^(x/3) =3^3 ⇒(x/3)=3 ⇒x=9 this is true, since 3^3 >1. but if ((x/3))^(x/3) =(0.8)^(0.8) ⇒(x/3)=0.8 ⇒x=2.4 this is not very true! since 1/(e)^(1/e) <0.8^(0.8) <1. the correct solution is: ((x/3))^(x/3) =(0.8)^(0.8) ⇒(x/3)=((0.8 ln 0.8)/(W(0.8 ln 0.8))) ⇒x=((2.4 ln 0.8)/(W(0.8 ln 0.8)))= { ((≈((2.4 ln 0.8)/(−2.72587187))=0.196467)),((≈((2.4 ln 0.8)/(−0.22314355))=2.4)) :}
$${if}\:\left(\frac{{x}}{\mathrm{3}}\right)^{\frac{{x}}{\mathrm{3}}} =\mathrm{3}^{\mathrm{3}} \:\Rightarrow\frac{{x}}{\mathrm{3}}=\mathrm{3}\:\Rightarrow{x}=\mathrm{9} \\ $$$${this}\:{is}\:{true},\:{since}\:\mathrm{3}^{\mathrm{3}} >\mathrm{1}. \\ $$$$ \\ $$$${but} \\ $$$${if}\:\left(\frac{{x}}{\mathrm{3}}\right)^{\frac{{x}}{\mathrm{3}}} =\left(\mathrm{0}.\mathrm{8}\right)^{\mathrm{0}.\mathrm{8}} \:\Rightarrow\frac{{x}}{\mathrm{3}}=\mathrm{0}.\mathrm{8}\:\Rightarrow{x}=\mathrm{2}.\mathrm{4} \\ $$$${this}\:{is}\:{not}\:{very}\:{true}!\:{since}\:\mathrm{1}/\sqrt[{{e}}]{{e}}<\mathrm{0}.\mathrm{8}^{\mathrm{0}.\mathrm{8}} <\mathrm{1}.\: \\ $$$${the}\:{correct}\:{solution}\:{is}: \\ $$$$\left(\frac{{x}}{\mathrm{3}}\right)^{\frac{{x}}{\mathrm{3}}} =\left(\mathrm{0}.\mathrm{8}\right)^{\mathrm{0}.\mathrm{8}} \: \\ $$$$\Rightarrow\frac{{x}}{\mathrm{3}}=\frac{\mathrm{0}.\mathrm{8}\:\mathrm{ln}\:\mathrm{0}.\mathrm{8}}{{W}\left(\mathrm{0}.\mathrm{8}\:\mathrm{ln}\:\mathrm{0}.\mathrm{8}\right)} \\ $$$$\Rightarrow{x}=\frac{\mathrm{2}.\mathrm{4}\:\mathrm{ln}\:\mathrm{0}.\mathrm{8}}{{W}\left(\mathrm{0}.\mathrm{8}\:\mathrm{ln}\:\mathrm{0}.\mathrm{8}\right)}=\begin{cases}{\approx\frac{\mathrm{2}.\mathrm{4}\:\mathrm{ln}\:\mathrm{0}.\mathrm{8}}{−\mathrm{2}.\mathrm{72587187}}=\mathrm{0}.\mathrm{196467}}\\{\approx\frac{\mathrm{2}.\mathrm{4}\:\mathrm{ln}\:\mathrm{0}.\mathrm{8}}{−\mathrm{0}.\mathrm{22314355}}=\mathrm{2}.\mathrm{4}}\end{cases} \\ $$
Commented by mr W last updated on 01/Apr/24
Commented by mr W last updated on 01/Apr/24
if x^x ≥1 ⇒there is only one solution for x. if (1/( (e)^(1/e) ))<x^x <1 ⇒there are two solutions for x. if x^x =(1/( (e)^(1/e) )) ⇒there is only one solution for x. if x^x <(1/( (e)^(1/e) )) ⇒there is no solution for x.
$${if}\:{x}^{{x}} \geqslant\mathrm{1}\:\Rightarrow{there}\:{is}\:{only}\:{one}\:{solution}\:{for}\:{x}. \\ $$$${if}\:\frac{\mathrm{1}}{\:\sqrt[{{e}}]{{e}}}<{x}^{{x}} <\mathrm{1}\:\Rightarrow{there}\:{are}\:{two}\:{solutions}\:{for}\:{x}. \\ $$$${if}\:{x}^{{x}} =\frac{\mathrm{1}}{\:\sqrt[{{e}}]{{e}}}\:\Rightarrow{there}\:{is}\:{only}\:{one}\:{solution}\:{for}\:{x}. \\ $$$${if}\:{x}^{{x}} <\frac{\mathrm{1}}{\:\sqrt[{{e}}]{{e}}}\:\Rightarrow{there}\:{is}\:{no}\:{solution}\:{for}\:{x}. \\ $$
Commented by Davidtim last updated on 01/Apr/24
thanks you are an exceptional of science. would you mind to prove the all of three forms?
$${thanks}\:{you}\:{are}\:{an}\:{exceptional}\:{of}\: \\ $$$${science}. \\ $$$${would}\:{you}\:{mind}\:{to}\:{prove}\:{the}\:{all}\:{of}\: \\ $$$${three}\:{forms}? \\ $$
Commented by mr W last updated on 01/Apr/24
y=x^x =e^(xln x) >0 y′=e^(xln x) (ln x+1)=x^x (ln x+1) y′=0: ln x+1=0 ⇒x=(1/e) that means: if x>(1/e),y′>0 ⇒y=x^x is strictly increasing if 0<x<(1/e),y′<0 ⇒y=x^x is strictly decreasing at x=(1/e), y_(min) =(1/( (e)^(1/e) ))
$${y}={x}^{{x}} ={e}^{{x}\mathrm{ln}\:{x}} >\mathrm{0} \\ $$$${y}'={e}^{{x}\mathrm{ln}\:{x}} \left(\mathrm{ln}\:{x}+\mathrm{1}\right)={x}^{{x}} \left(\mathrm{ln}\:{x}+\mathrm{1}\right) \\ $$$${y}'=\mathrm{0}:\:\mathrm{ln}\:{x}+\mathrm{1}=\mathrm{0}\:\Rightarrow{x}=\frac{\mathrm{1}}{{e}} \\ $$$${that}\:{means}: \\ $$$${if}\:{x}>\frac{\mathrm{1}}{{e}},{y}'>\mathrm{0}\:\Rightarrow{y}={x}^{{x}} \:{is}\:{strictly}\:{increasing} \\ $$$${if}\:\mathrm{0}<{x}<\frac{\mathrm{1}}{{e}},{y}'<\mathrm{0}\:\Rightarrow{y}={x}^{{x}} \:{is}\:{strictly}\:{decreasing} \\ $$$${at}\:{x}=\frac{\mathrm{1}}{{e}},\:{y}_{{min}} =\frac{\mathrm{1}}{\:\sqrt[{{e}}]{{e}}} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *