Show that $19^{31}>13^{33}$ 
How can i prove that $19^{31}>13^{33}$?

What I tried
$$\bigg(\frac{19}{13}\bigg)^2=\frac{361}{169}>2>1$$
then $19^{2}>13^{2}$ and $\displaystyle 19^{30}>13^{30}$
How do I show it. Help me please. 
 A: $$A=19^{31}\geq 19^{16}\cdot18^{15} = \color{red}{19^{16}\cdot 9^{15}} \cdot  2^{15}$$
$$B=13^{33} = 13\cdot 169^{16}\leq 13\cdot 171^{16} = 13 \cdot \color{red}{19^{16}\cdot 9^{15}}\cdot 9$$
Clearly  $$2^{15}\geq 13\cdot 9\implies A>B$$
A: Just compute the minimum $x$ such that $19^x > 13^{x+2}$. By taking logarithm on both sides, we have
$x \ln(19) > (x+2) \ln(13)$ 
and finally
$ x > 2 \frac{\ln(13)}{\ln(\frac{19}{13})}$ and the number on the right is smaller than 13 so way smaller than the case $x = 31$.
A: By your work
$$\frac{19^2}{13^2}>2,$$ which says
$$\left(\frac{19}{13}\right)^{31}>2^{15.5}>2^8=256>13^2$$ and we are done!
By this way we can prove that
$$19^{31}>13^{35}.$$
A: $19^{30}=(19^3)^{10}=6859^{10}\gt 13^{30}=(13^3)^{10}=2197^{10}$
$6859\gt 3\times 2197=6591$
$6859^{10}\gt (3\times 2197)^{10}=3^{10}\times 2197^{10}=3^3\times 3^3\times 3^3\times 3\times 2197^{10}=27^3\times 3\times 2197^{10}$
$27^3\times 3\times 2197^{10}\gt 13^3\times 2197^{10}$ already without even multiplying by $19.$
A: $\left[\dfrac{19}{13}\right]^{\large 4}\!\!\!=\!\left[1\!+\!\dfrac{6}{13}\right]^{\large 4}\!\!>1\!+\!4\left[\dfrac{6}{13}\right]\!>2\,\overset{(\ \ \,)^{\LARGE 8}}\Longrightarrow\, \left[\dfrac{19}{13}\right]^{\large 32}\!\!\!>2^{\large 8} > 19\cdot 13$
A: Note that by the binomial theorem
$$\left(\frac{19}{13}\right)^{31}=\left(1+\frac{6}{13}\right)^{31}>\binom{31}{3}\frac{6^3}{13^3}=\frac{31\cdot 30\cdot 29\cdot 36}{13^3}>\frac{26^4}{13^3}=16\cdot 13>13^2$$
which implies that $19^{31}>13^{33}$.
Bonus question. By using a similar approach we can show the stronger inequality $19^{31}>13^{34}$. Would you like to try it?
A: Well, using what you've got so far:
$(\frac {19}{13})^2 > 2$ so $(\frac {19}{13})^{30} > 2^{15}$  so $19^{30} > 2^{15}*13^{30}$.  
$19^{31} = 19^{30} *19 > 13^{30}*(19*2^{15}) {? \over >} 13^{30}*13^3 = 13^{33}$.  If we can show $19*(2^{15}) > 13^3$ we will be done.
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For what it's worth one way I might do it if it were up to me is to note.
$19 > 13$ so $19^k > 13^k$  For $19^k > 13^{k+2} = 13^k *13^2$ we need $(\frac {19}{13})^k > 169$.
$(\frac {19}{13})^2 > 2$ so $(\frac {19}{13})^{16} > 2^8 = 256> 169$.
So $19^{16} > 13^{18}$ and $19^{31} > 13^{33}$.... by a lot.
