How can I compare the numbers $2^{39}$, $5^{19}$ and $52^7$?

Hint: you can establish the right order between $5^{19}$ and $2^{39}$ as follows \begin{align} 5^{19} &= 5^{20-1}\\ &=\frac{5^{20}}{5}\\ &=\frac{(5^2)^{10}}{5}\\ &=\frac{25^{10}}{5}\\ &>\frac{16^{10}}{2}\\ &=\frac{(2^4)^{10}}{2}\\ &=2^{39} \end{align} then note that \begin{align} 52^7 &= 52^{10-3}\\ &=\frac{52^{10}}{52^3}\\ &=\frac{26^{10}\cdot 2^{10}}{26^{3}\cdot 2^{3}}\\ &<\frac{25^{10}\cdot 2^{10}}{26^{3}\cdot 2^{3}}\\ &<\frac{25^{10}\cdot 2^{10}}{26^{3}\cdot 5}\\ &=\frac{5^{20-1}\cdot 2^{10}}{26^{3}}\\ &=5^{19} \cdot \frac{2^{10}}{26^3}\\ &<5^{19} \end{align} and now you have to find if $52^7$ is somewhere in between $5^{19}$ and $2^{39}$ or under $2^{39}$.

First, we can easily compute some small powers manually to see the equalities in $$\phantom{(\ast)} \qquad 2^{11} = 2048 < 3^7 = 2187 < 5^5 = 3125. \qquad (\ast)$$

Multiplying both sides of the first inequality in $(\ast)$ by $2^{28} = 16^7$ gives the left-hand inequality in $$2^{39} < 3^7 \cdot 16^7 = 48^7 < 52^7 .$$ On the other hand, multiplying both sides of the second inequality in $(\ast)$ by $5^{14} = 25^7$ gives the right-hand inequality in $$52^7 < 3^7 \cdot 25^7 = 75^7 < 5^{19}.$$ Collecting the above inequalities gives: $$\color{#bf0000}{\boxed{2^{39} < 52^7 < 5^{19}}} .$$

To compare $2^{39}$ and $5^{19}$ we have

$$2^{39} = 2 \cdot 2^{38} = 2\cdot 4^{19} = 2 \cdot 4^4 \cdot 4^{15} = 512 \cdot4^{15} < 625 \cdot 5^{15} = 5^{4}\cdot 5^{15} = 5^{19}$$.