1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Intuition says that $4^{53}$ is larger. $\endgroup$ – jkabrg Jun 17 '15 at 18:54
  • 3
    $\begingroup$ Exponential growth is really fast. So even if the base is lesser by 1 , the fact that power is bigger by 9 matters more . $\endgroup$ – A Googler Jun 17 '15 at 18:56
  • 2
    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Jun 17 '15 at 18:59
  • 4
    $\begingroup$ Alternately, consider $\left(\frac54\right)^{44}$ vs. $4^9$... $\endgroup$ – abiessu Jun 17 '15 at 19:01
  • 2
    $\begingroup$ Also, $$5^{44} < 4^{53} \iff 10^{44} < 4^{75} = (4^5)^{15} = (1024)^{15}.$$ $\endgroup$ – Daniel Fischer Jun 17 '15 at 19:19
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The final step leading to the $\text{vs}$ comparison is to divide by $\left(\frac{5}{4}\right)^{18}$ a second time. $\endgroup$ – abiessu Jun 18 '15 at 2:21
8
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @YvesDaoust Thank you . $\endgroup$ – A Googler Jun 17 '15 at 19:11
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $102.164 \dots$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.