Is $202^{303}$ greater or $303^{202}$? 
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 
 A: 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!
A: $$\array{808 &\gt& 9 \\
101 \cdot 2^{3} &\gt& 3^{2} \\
101^{3} \cdot 2^{3} &\gt& 101^{2} \cdot 3^{2} \\
\left((101\cdot 2)^{3}\right)^{101} &\gt& \left((101\cdot 3)^{2}\right)^{101} \\
202^{303} &\gt& 303^{202}}$$
A: 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 $303^{202}<202^{303}$.
A: $\frac {202^{303}}{303^{202}}=\frac{2^{303}}{3^{202}}\frac{101^{303}}{101^{202}}=\frac{2}{3}^{202}*2^{101}*101^{101} = (4/3)^{202} \frac 12^{202}*2^{101}*101^{101} = (4/3)^{202}\frac 12^{101}*101^{101} = (4/3)^{202}*50.5^{101} > 1$
So $\frac {202^{303}}{303^{202}} > 1$  (by quite a LOT) 
So $202^{303} > 303^{202}$
A: Hint: The function $x \mapsto x^{1/x}$ has a single critical point at $x=e$ and is decreasing for $x \gt e$.
A: How to solve in your head: $202 > 2^7, 303 < 2^9$. Therefore $202^{303} > 2^{2121}, 303^{202} < 2^{1818}$.
Even easier: $202^{303} > 100^{303} = 1000^{202} > 303^{202}$.
A: $$
202 = 303^{\lg(202) / \lg(303)} \\
202^{303} = 303^{303 \lg(202)/ \lg(303)} \approx 303^{303\cdot11/12} > 303^{202}
$$
A: $$202^{303} = \left(\frac{2}{3}\right)^{303} \times 303^{303}= \left(\frac{2}{3}\right)^{303} \times 303^{101} \times 303^{202}$$
$$= \left(\left(\frac{2}{3}\right)^3\right)^{101}\times 303^{101}\times 303^{202}$$
$$=\left(\frac{8 \times 303}{27}\right)^{101} \times 303^{202}$$
$$=\left(\frac{808}{9}\right)^{101} \times 303^{202} > 303^{202}$$
A: $(202)^{303}=(2\times 101)^{3\times {101}}=2^{3\times 101} \times 101^{303}=8^{101}\times 101^{303}$
$(303)^{202} = (3 \times 101)^{2 \times 101} =  3^{2\times 101} \times 101^{202}=9^{101}\times 101^{202}$
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 $(202)^{303}>(303)^{202}$.
