Comparing two large numbers

Question

Can you compare two large exponential numbers, like $5^{44}$ and $4^{53}$ without taking their logs?

Answer 1

$$5^{44}<5^{45}=(5^3)^{15}=125^{15}<128^{15}=(256/2)^{15}=4^{60}/2^{15}<4^{53}$$ (because $2^{15}>2^{14}=4^{7}$)

Answer 2

Using a GCD-like approach, start by dividing out the "smaller" term:

$${5^{44}\over 4^{44}}=\left(\frac54\right)^{44}\\ {4^{53}\over 4^{44}}=4^9$$

Now the new "smaller" term is $\frac54$:

$${\left(\frac54\right)^{44}\over \left(\frac54\right)^{18}}=\left(\frac54\right)^{26}\\ {4^9\over \left(\frac54\right)^{18}}=\left(\frac85\right)^{18}\\ \left(\frac54\right)^{8}\text{ vs }\left(\frac{32}{25}\right)^{18}$$

By inspection, $\frac{32}{25}\ge\frac54$, and as the exponent on $\frac{32}{25}$ is also greater, it is the greater quantity. It is from the $4^{53}$ term, and therefore this is the greater original quantity.

Answer 3

One approach is to figure that, roughly, $2^{10} \approx 10^3$, and $5^9 \approx 2,000,000 = 2 \cdot 10^6$.

Then,

\begin{align} 4^{53} &= (2^2)^{53} \\ &= 2^{106} \\ &= 2^{100} \cdot 2^6 \\ &= (2^{10})^{10} \cdot 2^6 \\ &\approx (10^3)^{10} \cdot 2^6 \\ &= 10^{30} \cdot 2^6 \end{align}

By way of comparison,

\begin{align} 5^{44} &= 5^{45} \div 5 \\ &= (5^9)^5 \div 5 \\ &\approx (2 \cdot 10^6)^5 \div 5 \\ &= 2^5 \cdot (10^6)^5 \div 5 \\ &= 10^{30} \cdot (2^5 \div 5). \end{align}

Now, $2^6 > 2^5 \div 5$, so one might suppose $4^{53} > 5^{44}$.

Answer 4

I'll interpret the actual question as "without using a calculator?" since I assume that's your objection to using logs.

What I know about $5$ and $4$ is that $5^3 = 125$ is pretty close to but a bit smaller than $2^7 = 128$. That tells me that $\log_2 5 \le\frac{7}{3}$, hence that

$$\log_2 5^{44} \le \frac{308}{3} < 103$$

while $\log_2 4^{53} = 106$. So $4^{53}$ is bigger. Of course I could rephrase this argument without using logs by exponentiating everything but what's the point?

A calculator will verify that $\frac{7}{3}$ is in fact a convergent of $\log_2 5$, whose continued fraction approximation begins $2 + \frac{1}{3 + \frac{1}{9 + \dots}}$. So the above is a pretty close approximation; in fact the true value of $\log_2 5^{44}$ is $102.164 \dots$.

Answer 5

Binomial expansion can help.

$$ (x+y) = \sum_{k=0}^n\frac{n!}{k!(n-k)!}x^{n-k}y^k $$

In this case, x = 4, y = 1, which simplifies

For massive expansion like you have, you really only need to look at the first few terms (although with a low x and a big exponent, extra terms are advised. In this case:

$$ (4+1)^{44} \sim 4^{44} + 44\times4^{43} + \frac{44\times43}{2} \times 4^{42} + \frac{44\times43\times42}{6}\times 4^{41} + \frac{44\times43\times42\times41}{24}\times4^{40} + \frac{44\times43\times42\times41\times40}{120}\times4^{39} + \frac{44\times43\times42\times41\times40\times39}{720}\times4^{38} + \frac{44\times43\times42\times41\times40\times39\times38}{5040}\times4^{37} + \frac{44\times43\times42\times41\times40\times39\times38\times37}{40320}\times4^{36} + \frac{44\times43\times42\times41\times40\times39\times38\times37\times36}{362880}\times4^{35} + \frac{44\times43\times42\times41\times40\times39\times38\times37\times36\times35}{3628800}\times4^{34}$$

As the 4 terms get smaller, their relevance begin to diminish, and you can start to see the real relevance. So, $5^{44}$ is about $(1 + 11 + 50 + 170 + 416 + 833 + 1388 + 1984 + 2480 + 2755 + 2755 + ...) \times 4^{44} $ for that $5^{44}$. Adding them up, you see since $4^{53}/4^{44} = {4}^9 = 262144 $, which is WAY bigger than even a bunch of 2755s put together that $4^{53}$ is bigger.