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Question Number 205324 by hardmath last updated on 16/Mar/24

Compare: 37^(37) and 36^(38)

$$\mathrm{Compare}: \\ $$$$\mathrm{37}^{\mathrm{37}} \:\:\:\mathrm{and}\:\:\:\mathrm{36}^{\mathrm{38}} \\ $$

Answered by nikif99 last updated on 16/Mar/24

37^(37) ≶ 36^(38) ⇒ 37^(36) ×37 ≶ 36^(36) ×36^2 ⇒ ((37^(36) )/(36^(36) )) ≶ ((36^2 )/(37)) ⇒ (((37)/(36)))^(36) ≶ ((36^2 )/(37)) ⇒ (1+(1/(36)))^(36) ≶ ((36^2 )/(37)) ⇒ e < ((36^2 )/(37))

$$\mathrm{37}^{\mathrm{37}} \:\lessgtr\:\mathrm{36}^{\mathrm{38}} \:\Rightarrow \\ $$$$\mathrm{37}^{\mathrm{36}} ×\mathrm{37}\:\lessgtr\:\mathrm{36}^{\mathrm{36}} ×\mathrm{36}^{\mathrm{2}} \:\Rightarrow \\ $$$$\frac{\mathrm{37}^{\mathrm{36}} }{\mathrm{36}^{\mathrm{36}} }\:\lessgtr\:\frac{\mathrm{36}^{\mathrm{2}} }{\mathrm{37}}\:\Rightarrow \\ $$$$\left(\frac{\mathrm{37}}{\mathrm{36}}\right)^{\mathrm{36}} \:\lessgtr\:\frac{\mathrm{36}^{\mathrm{2}} }{\mathrm{37}}\:\Rightarrow \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{36}}\right)^{\mathrm{36}} \:\lessgtr\:\frac{\mathrm{36}^{\mathrm{2}} }{\mathrm{37}}\:\Rightarrow \\ $$$${e}\:<\:\frac{\mathrm{36}^{\mathrm{2}} }{\mathrm{37}} \\ $$

Answered by Frix last updated on 16/Mar/24

n∈N 1≤n≤4: n^n >(n−1)^(n+1) n≥5: n^n <(n−1)^(n+1)

$${n}\in\mathbb{N} \\ $$$$\mathrm{1}\leqslant{n}\leqslant\mathrm{4}:\:{n}^{{n}} >\left({n}−\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$$${n}\geqslant\mathrm{5}:\:{n}^{{n}} <\left({n}−\mathrm{1}\right)^{{n}+\mathrm{1}} \\ $$

Answered by Berbere last updated on 17/Mar/24

⇔ 37ln(37) and 38ln(36) x=36;claim (x+1)ln(x+1)≤(x+2)ln(x) ⇔(x+1)(ln(x)+ln(1+(1/x)))≤(x+2)ln(x) ⇔(x+1)ln(1+(1/x))≤ln(x) ln(1+a)=∫_0 ^a (1/(t+1))dt≤∫_0 ^a dt=a ⇒ln(1+(1/x))≤(1/x)⇒(x+1)ln(1+(1/x))≤(x+1).(1/x)=1+(1/x)≤2 ∀x≥1 ⇒1+(1/x)≤ln(x);∀x≥e^2 ;e^2 <36 ⇒37^(37) ≤36^(38)

$$\Leftrightarrow\:\mathrm{37}{ln}\left(\mathrm{37}\right)\:{and}\:\mathrm{38}{ln}\left(\mathrm{36}\right) \\ $$$${x}=\mathrm{36};{claim} \\ $$$$\left({x}+\mathrm{1}\right){ln}\left({x}+\mathrm{1}\right)\leqslant\left({x}+\mathrm{2}\right){ln}\left({x}\right) \\ $$$$\Leftrightarrow\left({x}+\mathrm{1}\right)\left({ln}\left({x}\right)+{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\right)\leqslant\left({x}+\mathrm{2}\right){ln}\left({x}\right) \\ $$$$\Leftrightarrow\left({x}+\mathrm{1}\right){ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\leqslant{ln}\left({x}\right) \\ $$$${ln}\left(\mathrm{1}+{a}\right)=\int_{\mathrm{0}} ^{{a}} \frac{\mathrm{1}}{{t}+\mathrm{1}}{dt}\leqslant\int_{\mathrm{0}} ^{{a}} {dt}={a} \\ $$$$\Rightarrow{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\leqslant\frac{\mathrm{1}}{{x}}\Rightarrow\left({x}+\mathrm{1}\right){ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\leqslant\left({x}+\mathrm{1}\right).\frac{\mathrm{1}}{{x}}=\mathrm{1}+\frac{\mathrm{1}}{{x}}\leqslant\mathrm{2} \\ $$$$\forall{x}\geqslant\mathrm{1}\:\Rightarrow\mathrm{1}+\frac{\mathrm{1}}{{x}}\leqslant{ln}\left({x}\right);\forall{x}\geqslant{e}^{\mathrm{2}} ;{e}^{\mathrm{2}} <\mathrm{36} \\ $$$$\Rightarrow\mathrm{37}^{\mathrm{37}} \leqslant\mathrm{36}^{\mathrm{38}} \\ $$$$ \\ $$

Commented by hardmath last updated on 18/Mar/24

cool dear professors thank you

$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professors}\:\mathrm{thank}\:\mathrm{you} \\ $$

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