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Question Number 172323 by Giantyusuf last updated on 25/Jun/22
which number is greater  22^(55)  and 55^(22)  ???
$${which}\:{number}\:{is}\:{greater} \\ $$$$\mathrm{22}^{\mathrm{55}} \:\boldsymbol{{and}}\:\mathrm{55}^{\mathrm{22}} \:??? \\ $$
Answered by nurtani last updated on 25/Jun/22
22^(55)  > 55^(22)
$$\mathrm{22}^{\mathrm{55}} \:>\:\mathrm{55}^{\mathrm{22}} \\ $$
Answered by floor(10²Eta[1]) last updated on 25/Jun/22
2^(55) .11^(55) >5^(22) .11^(22)   2^(55) .11^(33) >5^(22)   clearly true since 11^(33) >5^(33) >5^(22)   ⇒22^(55) >55^(22)
$$\mathrm{2}^{\mathrm{55}} .\mathrm{11}^{\mathrm{55}} >\mathrm{5}^{\mathrm{22}} .\mathrm{11}^{\mathrm{22}} \\ $$$$\mathrm{2}^{\mathrm{55}} .\mathrm{11}^{\mathrm{33}} >\mathrm{5}^{\mathrm{22}} \\ $$$$\mathrm{clearly}\:\mathrm{true}\:\mathrm{since}\:\mathrm{11}^{\mathrm{33}} >\mathrm{5}^{\mathrm{33}} >\mathrm{5}^{\mathrm{22}} \\ $$$$\Rightarrow\mathrm{22}^{\mathrm{55}} >\mathrm{55}^{\mathrm{22}} \\ $$
Answered by Gazella thomsonii last updated on 25/Jun/22
  A=22^(55)   B=55^(22)   lnA=55ln(22)  ln(B)=22ln(55)  ln(22)≈3.091  ln(55)≈4.007  lnA=165.xxx  lnB=88.xxx  ∴ B<A
$$ \\ $$$$\mathrm{A}=\mathrm{22}^{\mathrm{55}} \:\:\mathrm{B}=\mathrm{55}^{\mathrm{22}} \\ $$$$\mathrm{lnA}=\mathrm{55ln}\left(\mathrm{22}\right)\:\:\mathrm{ln}\left(\mathrm{B}\right)=\mathrm{22ln}\left(\mathrm{55}\right) \\ $$$$\mathrm{ln}\left(\mathrm{22}\right)\approx\mathrm{3}.\mathrm{091}\:\:\mathrm{ln}\left(\mathrm{55}\right)\approx\mathrm{4}.\mathrm{007} \\ $$$$\mathrm{lnA}=\mathrm{165}.\mathrm{xxx}\:\:\mathrm{lnB}=\mathrm{88}.\mathrm{xxx} \\ $$$$\therefore\:\mathrm{B}<\mathrm{A} \\ $$
Answered by MJS_new last updated on 25/Jun/22
f(x)=x^(1/x)  has its maximum at x=e ⇒  ∀a,b∣e<a<b: a^(1/a) >b^(1/b)   ⇔  ((ln a)/a)>((ln b)/b)  ⇔  bln a >aln b  ⇔  a^b >b^a     for a ,b ∈N∧a, b>1 the only exceptions are  2^3 <3^2   2^4 =4^2
$${f}\left({x}\right)={x}^{\frac{\mathrm{1}}{{x}}} \:\mathrm{has}\:\mathrm{its}\:\mathrm{maximum}\:\mathrm{at}\:{x}=\mathrm{e}\:\Rightarrow \\ $$$$\forall{a},{b}\mid\mathrm{e}<{a}<{b}:\:{a}^{\frac{\mathrm{1}}{{a}}} >{b}^{\frac{\mathrm{1}}{{b}}} \\ $$$$\Leftrightarrow \\ $$$$\frac{\mathrm{ln}\:{a}}{{a}}>\frac{\mathrm{ln}\:{b}}{{b}} \\ $$$$\Leftrightarrow \\ $$$${b}\mathrm{ln}\:{a}\:>{a}\mathrm{ln}\:{b} \\ $$$$\Leftrightarrow \\ $$$${a}^{{b}} >{b}^{{a}} \\ $$$$ \\ $$$$\mathrm{for}\:{a}\:,{b}\:\in\mathbb{N}\wedge{a},\:{b}>\mathrm{1}\:\mathrm{the}\:\mathrm{only}\:\mathrm{exceptions}\:\mathrm{are} \\ $$$$\mathrm{2}^{\mathrm{3}} <\mathrm{3}^{\mathrm{2}} \\ $$$$\mathrm{2}^{\mathrm{4}} =\mathrm{4}^{\mathrm{2}} \\ $$

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