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

Question

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.

Answer 1

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: $$\color{#bf0000}{\boxed{2^{39} < 52^7 < 5^{19}}} .$$

Answer 2

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 $2^{39}$.

Answer 3

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

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

Answer 4

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

$$5^{19}>52^7>2^{39}. $$

Answer 5

$$ 52^7 = (26^7 \cdot 2^7) > ((2^{4.7})^7 \cdot 2^7) = 2^{39.9} > 2^{39} $$

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,

$$ 2^{39} < 52^7 < 5^{19}$$