Comparing two large numbers

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

• Intuition says that $4^{53}$ is larger. – jkabrg Jun 17 '15 at 18:54
• Exponential growth is really fast. So even if the base is lesser by 1 , the fact that power is bigger by 9 matters more . – A Googler Jun 17 '15 at 18:56
• Would you accept a mental arithmetic calculation using logs (base 10)? log 2=0.301, so log 5=1-log2 <0.7, so 44*log 5<30.8. OTOH log 4=2 log 2>0.6, so 53*log 4>31.8. – Jyrki Lahtonen Jun 17 '15 at 18:59
• Alternately, consider $\left(\frac54\right)^{44}$ vs. $4^9$... – abiessu Jun 17 '15 at 19:01
• Also, $$5^{44} < 4^{53} \iff 10^{44} < 4^{75} = (4^5)^{15} = (1024)^{15}.$$ – Daniel Fischer Jun 17 '15 at 19:19

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.

• The final step leading to the $\text{vs}$ comparison is to divide by $\left(\frac{5}{4}\right)^{18}$ a second time. – abiessu Jun 18 '15 at 2:21

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

• @YvesDaoust Thank you . – A Googler Jun 17 '15 at 19:11

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 $4^{53} > 5^{44}$.

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 $102.164 \dots$.

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:

$$(4+1)^{44} \sim 4^{44} + 44*4^{43} + \frac{44*43}{2} * 4^{42} + \frac{44*43*42}{6}* 4^{41} + \frac{44*43*42*41}{24}*4^{40} + \frac{44*43*42*41*40}{120}*4^{39} + \frac{44*43*42*41*40*39}{720}*4^{38} + \frac{44*43*42*41*40*39*38}{5040}*4^{37} + \frac{44*43*42*41*40*39*38*37}{40320}*4^{36} + \frac{44*43*42*41*40*39*38*37*36}{362880}*4^{35} + \frac{44*43*42*41*40*39*38*37*36*35}{3628800}*4^{34}$$

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 $(1 + 11 + 50 + 170 + 416 + 833 + 1388 + 1984 + 2480 + 2755 + 2755 + ...) * 4^{44}$ 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.