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Find without use of calculator which of the two numbers is greater $202^{303}$ or $303^{202}$.
I think we have to do this with calculus because I got this question from my calculus book.
I tried searching on google and SE but didn't get required solution
Michael Lugo
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asked May 15, 2016 at 21:35
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$$\array{808 &\gt& 9 \\ 101 \cdot 2^{3} &\gt& 3^{2} \\ 101^{3} \cdot 2^{3} &\gt& 101^{2} \cdot 3^{2} \\ \left[[101\cdot 2]{3}\right]{101} &\gt& \left[[101\cdot 3]{2}\right]{101} \\ 202^{303} &\gt& 303^{202}}$$
miracle173
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answered May 15, 2016 at 22:03
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We have $a^b 1$ [by quite a LOT]
So $202^{303} > 303^{202}$
answered May 15, 2016 at 23:46
fleabloodfleablood
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Hint: The function $x \mapsto x^{1/x}$ has a single critical point at $x=e$ and is decreasing for $x \gt e$.
answered May 16, 2016 at 0:28
lhflhf
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How to solve in your head: $202 > 2^7, 303 < 2^9$. Therefore $202^{303} > 2^{2121}, 303^{202} < 2^{1818}$.
Even easier: $202^{303} > 100^{303} = 1000^{202} > 303^{202}$.
Umberto
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answered May 16, 2016 at 11:07
gnasher729gnasher729
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$$ 202 = 303^{\lg[202] / \lg[303]} \\ 202^{303} = 303^{303 \lg[202]/ \lg[303]} \approx 303^{303\cdot11/12} > 303^{202} $$
answered May 16, 2016 at 12:46
Olba12Olba12
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$$202^{303} = \left[\frac{2}{3}\right]{303} \times 303{303}= \left[\frac{2}{3}\right]{303} \times 303{101} \times 303^{202}$$ $$= \left[\left[\frac{2}{3}\right]3\right]{101}\times 303^{101}\times 303^{202}$$ $$=\left[\frac{8 \times 303}{27}\right]{101} \times 303{202}$$ $$=\left[\frac{808}{9}\right]{101} \times 303{202} > 303^{202}$$
answered May 9, 2018 at 20:39
John_dydxJohn_dydx
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$[202]{303}=[2\times 101]{3\times {101}}=2^{3\times 101} \times 101^{303}=8^{101}\times 101^{303}$
$[303]{202} = [3 \times 101]{2 \times 101} = 3^{2\times 101} \times 101^{202}=9^{101}\times 101^{202}$
So $$\frac{[202]{303}}{[303]{202}}= \frac{8^{101}\times 101^{303}}{9^{101}\times 101^{202}}= \frac{8^{101}}{9^{101}}\times 101^{101}=\left[\frac{808}{9}\right]^{101}$$