if a^a =b^b ; a=b is it true? if it is true then prove it.


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Question Number 205849 by Davidtim last updated on 31/Mar/24

$i f a^{a} = b^{b}; a = b i s i t t r u e ?$ $i f i t i s t r u e t h e n p r o v e i t .$
Commented by mr W last updated on 31/Mar/24
$not true for 0<a, b<1. example: a, b=((ln 0.8)/(W(ln 0.8)))≈ { ((0.094649710865)),((0.739533650011)) :} 0.094649710865^(0.094649710865) =0.739533650011^(0.739533650011) =0.8$
$n o t t r u e f o r 0 < a, b < 1 .$ $e x a m p l e :$ $a, b = \frac{\ln 0.8}{W (\ln 0.8)} \approx {\begin{cases} 0.094649710865 \\ 0.739533650011 \end{cases}$ $0 . 094649710865^{0.094649710865}$ $= 0 . 739533650011^{0.739533650011}$ $= 0.8$
Commented by Davidtim last updated on 01/Apr/24

$p l e a s e c h e c k t h e a b o v e e x p .$
Commented by Davidtim last updated on 01/Apr/24

Commented by mr W last updated on 01/Apr/24
$if ((x/3))^(x/3) =3^3 ⇒(x/3)=3 ⇒x=9 this is true, since 3^3 >1. but if ((x/3))^(x/3) =(0.8)^(0.8) ⇒(x/3)=0.8 ⇒x=2.4 this is not very true! since 1/(e)^(1/e) <0.8^(0.8) <1. the correct solution is: ((x/3))^(x/3) =(0.8)^(0.8) ⇒(x/3)=((0.8 ln 0.8)/(W(0.8 ln 0.8))) ⇒x=((2.4 ln 0.8)/(W(0.8 ln 0.8)))= { ((≈((2.4 ln 0.8)/(−2.72587187))=0.196467)),((≈((2.4 ln 0.8)/(−0.22314355))=2.4)) :}$
$i f {(\frac{x}{3})}^{\frac{x}{3}} = 3^{3} \Rightarrow \frac{x}{3} = 3 \Rightarrow x = 9$ $t h i s i s t r u e, s i n c e 3^{3} > 1 .$ $b u t$ $i f {(\frac{x}{3})}^{\frac{x}{3}} = {(0.8)}^{0.8} \Rightarrow \frac{x}{3} = 0.8 \Rightarrow x = 2.4$ $t h i s i s n o t v e r y t r u e! s i n c e 1 / \sqrt[e]{e} < 0 . 8^{0.8} < 1 .$ $t h e c o r r e c t s o l u t i o n i s :$ ${(\frac{x}{3})}^{\frac{x}{3}} = {(0.8)}^{0.8}$ $\Rightarrow \frac{x}{3} = \frac{0.8 \ln 0.8}{W (0.8 \ln 0.8)}$ $\Rightarrow x = \frac{2.4 \ln 0.8}{W (0.8 \ln 0.8)} = {\begin{cases} \approx \frac{2.4 \ln 0.8}{- 2.72587187} = 0.196467 \\ \approx \frac{2.4 \ln 0.8}{- 0.22314355} = 2.4 \end{cases}$
Commented by mr W last updated on 01/Apr/24

Commented by mr W last updated on 01/Apr/24

$i f x^{x} ⩾ 1 \Rightarrow t h e r e i s o n l y o n e s o l u t i o n f o r x .$ $i f \frac{1}{\sqrt[e]{e}} < x^{x} < 1 \Rightarrow t h e r e a r e t w o s o l u t i o n s f o r x .$ $i f x^{x} = \frac{1}{\sqrt[e]{e}} \Rightarrow t h e r e i s o n l y o n e s o l u t i o n f o r x .$ $i f x^{x} < \frac{1}{\sqrt[e]{e}} \Rightarrow t h e r e i s n o s o l u t i o n f o r x .$
Commented by Davidtim last updated on 01/Apr/24

$t h a n k s y o u a r e a n e x c e p t i o n a l o f$ $s c i e n c e .$ $w o u l d y o u m i n d t o p r o v e t h e a l l o f$ $t h r e e f o r m s ?$
Commented by mr W last updated on 01/Apr/24

$y = x^{x} = e^{x \ln x} > 0$ $y^{'} = e^{x \ln x} (\ln x + 1) = x^{x} (\ln x + 1)$ $y^{'} = 0 : \ln x + 1 = 0 \Rightarrow x = \frac{1}{e}$ $t h a t m e a n s :$ $i f x > \frac{1}{e}, y^{'} > 0 \Rightarrow y = x^{x} i s s t r i c t l y i n c r e a s i n g$ $i f 0 < x < \frac{1}{e}, y^{'} < 0 \Rightarrow y = x^{x} i s s t r i c t l y d e c r e a s i n g$ $a t x = \frac{1}{e}, y_{m i n} = \frac{1}{\sqrt[e]{e}}$
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