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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

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    $\begingroup$ Did you try anything more than just searching the Internet? $\endgroup$ – polettix May 15 '16 at 21:41
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    $\begingroup$ That would be also my question. But just to give you a hint, do you know the properties of $log$? $\endgroup$ – Umberto May 15 '16 at 21:42
  • $\begingroup$ $$202^{303} \lt 303^{202} $$ $$\Leftrightarrow \sqrt[202]{202} \lt \sqrt[303]{303}$$ $$\Leftrightarrow \frac{202}{\log 202} < \frac{303}{\log 303}$$ $\endgroup$ – miracle173 May 15 '16 at 21:46
  • $\begingroup$ Huh......for instance i just forgot about there is a thing called $\log$ $\endgroup$ – user5954246 May 15 '16 at 21:48
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    $\begingroup$ This might be useful: Fastest way to check if $x^y > y^x$? and [Given $a>b>2$ both positive integers, which of $a^b$ and $b^a$ is larger?] (math.stackexchange.com/questions/410697/…) (and you can probably find several other similar questions if you look at the post listed among related question in the sidebar). $\endgroup$ – Martin Sleziak May 16 '16 at 13:25
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We have $a^b<b^a$ iff $b\log(a)<a\log(b)$ iff $\frac{\log(a)}a<\frac{\log(b)}b$. Consider the function: $$f:x\mapsto\frac{\log(x)}x:\mathbb R^+\to\mathbb R$$ Then: $$f'(x)=\frac{1-\log(x)}{x^2}$$ Hence $$f'(x)>0\quad\mathrm{iff}\quad x<e$$ Hence $f$ is decreasing for $x>e$ and this proves $f(303)<f(202)$, hence $303^{202}<202^{303}$.

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$$\array{808 &\gt& 9 \\ 101 \cdot 2^{3} &\gt& 3^{2} \\ 101^{3} \cdot 2^{3} &\gt& 101^{2} \cdot 3^{2} \\ \left((101\cdot 2)^{3}\right)^{101} &\gt& \left((101\cdot 3)^{2}\right)^{101} \\ 202^{303} &\gt& 303^{202}}$$

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We are comparing $202^{303}$ and $303^{202}$.

$202^{303}$ is equal to $202^{202}$ * $202^{101}$.

$303^{202}$ is equal to $(202 * 1.5)^{202}$ which is equal to $202^{202}$ * $1.5^{202}$

Now, we can divide out the $202^{202}$ from both sides which yields $202^{101}$ versus $1.5^{202}$. $1.5^{202}$ can be written as $2.25^{101}$ (squaring the inside, thus dividing the exponent by 2). Since $202^{101}$ > $2.25^{101}$, $202^{303}$ > $303^{202}$. No need for calculus!

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$\frac {202^{303}}{303^{202}}=\frac{2^{303}}{3^{202}}\frac{101^{303}}{101^{202}}=\frac{2}{3}^{202}*2^{101}*101^{101} = (4/3)^{202} \frac 12^{202}*2^{101}*101^{101} = (4/3)^{202}\frac 12^{101}*101^{101} = (4/3)^{202}*50.5^{101} > 1$

So $\frac {202^{303}}{303^{202}} > 1$ (by quite a LOT)

So $202^{303} > 303^{202}$

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Hint: The function $x \mapsto x^{1/x}$ has a single critical point at $x=e$ and is decreasing for $x \gt e$.

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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: $202^{303} > 100^{303} = 1000^{202} > 303^{202}$.

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$$ 202 = 303^{\lg(202) / \lg(303)} \\ 202^{303} = 303^{303 \lg(202)/ \lg(303)} \approx 303^{303\cdot11/12} > 303^{202} $$

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$$202^{303} = \left(\frac{2}{3}\right)^{303} \times 303^{303}= \left(\frac{2}{3}\right)^{303} \times 303^{101} \times 303^{202}$$ $$= \left(\left(\frac{2}{3}\right)^3\right)^{101}\times 303^{101}\times 303^{202}$$ $$=\left(\frac{8 \times 303}{27}\right)^{101} \times 303^{202}$$ $$=\left(\frac{808}{9}\right)^{101} \times 303^{202} > 303^{202}$$

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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 $$\frac{(202)^{303}}{(303)^{202}}= \frac{8^{101}\times 101^{303}}{9^{101}\times 101^{202}}= \frac{8^{101}}{9^{101}}\times 101^{101}=\left(\frac{808}{9}\right)^{101}$$ As $808>9$ we have $(202)^{303}>(303)^{202}$.

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