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Which one is larger: $$37^{38}\quad\text{or}\quad38^{37}$$

I solved it as follow:

First I divided both numbers by $37^{37}$ to get $37$ and $(\frac{38}{37})^{37}$. Since $(1+\frac1{37})^{37}<e<37$ We have $$37^{38}>38^{37}$$ I'm looking for other approaches. Can you please solve it differently?

J. W. Tanner
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Etemon
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  • With a slide rule, $$38\log 37 = 38 + 38\log 3.7 = 38 + 38\times 0.568 = 38+21.6=59.6,$$ and $$37\log 38 = 37+37\log 3.8 = 37+37 \times 0.58 \approx 37+21.45=58.45,$$so $37^{38}>38^{37}$. In fact, the former is more than an order of magnitude bigger, like you calculated. – Benjamin Wang Apr 28 '22 at 10:24

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