How can I compare the numbers $2^{39}$, $5^{19}$ and $52^7$? I have to compare the numbers $2^{39}$, $5^{19}$ and $52^7$. I don't know how to do that because their exponents don't have anything in common. 
 A: 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 $2^{39}$.
A: 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:
$$\color{#bf0000}{\boxed{2^{39} < 52^7 < 5^{19}}} .$$
A: To compare $2^{39}$ and $5^{19}$ we have
$$2^{39} = 2 \cdot 2^{38} = 2\cdot 4^{19} = 2 \cdot 4^4 \cdot 4^{15} = 512 \cdot4^{15} < 625 \cdot 5^{15} = 5^{4}\cdot 5^{15} = 5^{19}$$.
A: Since $5=2^{\log _2 5}$ we have
$$5^{19}=2^{19 \log_2 5}.$$
Now we get some estimates on that logarithm:
We have $2^{2.25}=4 \sqrt[4]{2}<5$, as $\sqrt[4]{2} < \frac{5}{4}=1.25$. This means $\log_2 5>2.25$ and thus
$$5^{19}=2^{19 \log_2 5}> 2^{19 \times 2.25}=2^{42.75}>2^{39}.$$
Also,
$$52^7=2^{7 \log_2 52} $$
and since $2^{5 \frac{2}{3}}=\frac{64}{\sqrt[3]{2}}<52$ we have
$$52^7=2^{7 \log_2 52}>2^{7\times 5 \frac{2}{3}}=2^{39 \frac{2}{3}}>2^{39}. $$
In summary, we have
$$5^{19}>52^7>2^{39}. $$
A: $$ 52^7 = (26^7 \cdot 2^7) > ((2^{4.7})^7 \cdot 2^7) = 2^{39.9} > 2^{39} $$
As $26 = 2^{\frac{\ln 26}{ \ln 2}} \approx 2^{4.7004}$.
Also,
$$ 52^7 < 6^7 \cdot 10^7 = 6^7 5^7 2^7 < 6^7 5^7 5^4 = 5^{7 \cdot \frac{\ln 6}{\ln 5}} 5^7 5^4 = 5^{7.793 + 7 + 4} < 5^{19}$$
Using $2^7 < 5^4$.
Thus,
$$ 2^{39} < 52^7 < 5^{19}$$
