Which Is Larger? | Comparison: 40^39 and 39^40 | Can you think up this method?
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- čas přidán 24. 02. 2023
- Comparison, which one is greater? A fantastic math Problem. Watch the video and find out the answer!
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Take the logarithm of both numbers 40^39 and 39^40 : you then have to compare 39log40 and 40log39, or log40/40 and log39/39, which boils down to study the growth of the function logx/x (evaluated at 40 and 39). The derivative of logx/x is 1/x .1/x - logx . 1/x² = 1/x² . (1-logx) which is negative at 39 or 40. Hence, the inequality log40/40 < log39/39, and at last 40^39 < 39^40
Yeh that's what I initially did
Imdid this
I did the same 😊. It is clear that whenever the numbers (f.ex. 39 and 40) are greater than e, the same inequality holds, because logx/x is decreasing. If both numbers were smaller than e, the opposite inequality holds.
I did almost exactly the same. But all iny head, simply graphing both the id and log functions in my head and noting that the growth between any two integers of the id function will always be bigger than the growth of the log function for x>1.
yea but this question is meant for 7 - 8 graders who dont know calculus
you can also use (1 + 1/n)^n < 3 (it comes from the definition of "e"). So you can evaluate (1+1/39)^39 / 39 < 3/39 = 1/13 < 1
e = 3?🥲🥲🥲🥲
x^(1/x) gets smaller for x>e, and since e40^39
Solve in the general case, compare a^b vs b^a where a, b is real and a, b >= e
a^b vs b^a
(a^b)^(1/bc) vs (b^a)^(1/bc)
a^(1/a) vs b^(1/b)
analize function y=x^(1/x).
Differentiate y(x):
y' = -x^(x-2) * (log(x) - 1)
for x > e: y'(x) < 0, therefore if a > b >= e then a^(1/a) < b^(1/b) => a^b < b^a
In particular, 40^39 < 39^40, or e^Pi > Pi^e
(a^b)^(1/bc) vs (b^a)^(1/bc) should be (a^b)^[1/(bc)] vs (b^a)^[1/(bc)] because of the Order of Operations. You need grouping symbols in the denominators.
You can write 40^39 as (39+1)^39, and use the binomian formula. You get 40 summands, where the first 2 summands are 39^39, and the 38 remaining are each much smaller than 0.5*39^39 - so 40^39 is smaller than 39*39^39.
Or you could also write 39 as 40-1.
Same thing.
youtube.com/@Learninghub21
a short solution would be that (1 + 1/39)^39 = e and because e/39 < 1 40^39 is smaller than 40^39
That's an intuitive approach to it, but e is a transcendental number that can't be a product of nontrancendental numbers such as 39 🤓
@@axelpacheco7470 Strictly correct is (1 + 1/39)^39 < e, and dem Hauke's logic holds.
(1 + 1/n)^n -> e as n -> infty, and 1 < (1 + 1/n)^n < e for finite positive n.
Isnt there a need for limit tends to ♾️ for getting e as answer...Please explain kindly
@@wasalawyer.1179 the function (1 + 1/n)^n increases as n increases, which can be proved by a few different methods, converging to e at the limit as n goes to infinity, thus you can argue that the value of (1 + 1/n)^n is less than e for any arbitrary value of n.
It is not e 😢
I'd use calculus to solve this problem, but your approach is quite interesting. Thank you!
ارجو ان توضح لنا طريقتك حتى نستفيد منها..
This reminds me the old problem of demonstrating that e^pi > pi^e (of course, without making explicit computations). However, much more interesting is the general problem of comparing a^b with b^a.
Good job. Clear and easy to understand.
do you have a 6 hour version of this?
You can try out small values and see that after 3^2 and 2^3 the greater exponent proves larger.
@@hxqing ???? IF n=4, 4^3 < 3^4
@@pollen_allergy Right, the expression with the larger exponent is larger in value.
Yeah, and then you can prove that if you have (a^(a-1) < (a-1)^a) for some small a, then each step up the inequality just gets bigger... It is all just recursion.
@ A. Hardin -- That is false, as a counter example is shown below.
@@hxqing -- You wrote that wrong, because you are missing grouping symbols: n^(n - 1) > (n - 1)^n, for appropriate n-values.
I like how others approached this more. I brute forced directly from the original numbers and performed derivative on x^(79-x). It got me to the right answer, but the formulation of other people just seemed better
A better approach might be to take the derivative of x^(1560/x). This is the 1560th power of x^(1/x), which has a global maximum at x=e and decreases afterward. This same method lets you compare a^b versus b^a whenever a and b are both at least e: the one with the smaller base is larger
"brute force" bawahaha
I inspired from your CZcams video about which is larger and from your idea, i found the approximation of π correct to 6 decimal places by suitable multiplication of 40^39/39^40 by the number consists of integer and fractional part sir. Thanks for your effort.
Just notice that (1+1/n)^n < e, for any n E |N. And e < 39, hence e/39 < 1.
Beautiful solution. Thanks for sharing
You are a genius to complicate it. I got a much simple way from many comments.
So, generally, we have the question: a^b and b^a, which is larger? Consider first a>0 and b>0 and a1, has a minimum in x = e, and this minimum is exactly y = e (in effect, y = e/loge = e/1 = e)
- hence this function is decreasing with x < e and increasing with x > e
- hence, with a < e and b < e , we have that a^b < b^a, while with a > e and b > e, we have that a^b > b^a (hence, with a=39 and b=40 we have that 39^40>40^39)
- interesting enough, because the function x/logx has a minimum in x=e, if ae, it is possible that a^b = b^a. For this, it is necessary to solve the equation x/logx = w, where there are two solutions w1 and w2, ie a=w1 and b=w2, by which a^b=b^a exactly.
If e40^39....
If you write 40^39 as (39+1)^39,and make (a+b)^n,binimoial expression as a^n+a^(n-1)*b*(n;1)++a^(n-2)*b^2*(n;2).....+b^n, write a as 39,and b as 1,then you will notice combination cofficients,39 th power parameters product( are after 3rd term) less than 39^39. For sum of 40 terms in expression for 40^39 is less than 39^40.
That's how I did it too.
Thank you for showing this interesting and smart solution.
As a programmer
var a=40**39;
var b = 39**40;
if (a>b){
console.log("a is greater");
}
else{
console.log("b is greater");
}
Results is B is greater means 39**40 is greater;; 🙂
It can be noted that if x^(1/X) = A, and A is larger than 1 and less than e^(1/e), then the equation has 2 solutions. For example 2^(1/2) = 4^(1/4). And approximately for example 1.0703^(1/1.0703) = 66^(1/66). Sketching the graph y=x^(1/X) Will make this clear.
Usually in problems like this, the larger exponent wins. The real question is when does x^(x-1)=(x-1)^x. I believe it occurs at a low number.
When you got to (1+1/39)^39 * 1/39, I immediately thought about how (1+1/x)^x goes to e, as x increases. As such, e/39 < 1, so 39^40 is bigger.
Did that method too!
Same as my solution ^^
Interesting method. You can simplify it if you recognize that lim n->inf (1 + 1/n)^n = e and (1 + 1/2)^2 = 9/4 both of which are less than 39 meaning that (1 + 1/39)^39 must be less than 39 since it exists between them so the equation must be less than 1.
After division when we make the powers same. We should cancel the power 39 from numerator and denominator. Now the left expression will be 40/39.39 which means the answer will be smaller than 1. Hence proved that 40^39 < 39^40
Any number with base a (a greater than or equal to 3) and an exponent of a+1 must be greater than a number with base a+1 and an exponent of A. For example: 5 to the sixth power must be greater than 6 to the fifth power; 18 to the 19th power must be greater than 19 to the 18th ------, 39 to the 40th power must be greater than 40 to the 39th power, and so on, without exception. In fact, this can be used as a mathematical formula, no longer need to do tedious calculations to prove who is big and who is small. If you want to prove that this formula works, then prove that this formula is true. You don't have to calculate the magnitude of every two numbers.
Thanks for the explanation! Pretty much I used the same method 👍
Simple:
40³⁹ Vs 39⁴⁰ ( Taking Logarithm to base 40 for both sides)
39Log₄₀40 Vs 40Log₄₀39
39x1 Vs 40x 0.9931
39 Vs 39.72
39 < 39.72
This method maybe the fastest method to solve this math
Very good proof. Knowing a fair bit of calculus one knows that if the sum of the base and exponent is equal and larger than 2e, then the number with the larger exponent is larger. 79 is clearly larger than 2e. Of course it requires the base to be at least e in general. I mean: A is at least e (Eulers number) and B is larger than A, then A^B is larger than B^A.
This is *almost* a valid proof for all sequential pairs. There are 3 exceptions. These are the pairs 0&1, 1&2 and 2&3. In the first case either arrangement equals 0. In the second and third cases, putting the smaller number as the exponent actually results in the larger value.
Yes, almost. 1^0 = 1. Yes, because 2+3 is less than 2e. See my comments and thank you four your comment.
Suppose 39⁴⁰ ≤ 40⁹
⇔ 39 ≤ (40/39)³⁹ = (1 + 1/39)³⁹ < e < 3,
which is a contradiction, so we conclude that 39⁴⁰ > 40⁹.
Whenever e40^39
For n=m+1, n>3, and m>2, then
n^m will always be less than m^n
amazing video! i love how you strictly used algebra. yes u could all these other ways in the comments but using just algebra is more clever in my opinion
Apply log on both side.. 40log39 is greater than 39log40
*edit, seems like many mention this. And ofc this is not really abstract.
No need to do this.
Just start with the low ones, discover the trend, draw conclusion.
3^2 2^3 = 9 vs 8 = 0.89
4^3 3^4 = 64 vs 81 = 1,27
5^4 4^5 = 625 vs 1025 = 1,64
6^5 5^6 = 7776 vs 15625 = 20,14
40 squared 39 rooted 40 is 36
39 squared 40 rooted 39 is 42
= the last multiplication (40th) makes the number always bigger if the base values are similar/sequential
you could have used bernoulli's inequality on (1+1/39)^39. As 1/39
x^y when x is close in size to x!, and y < y! will be consistent
eq1: y=x^(x+1)-(x+1)^(x)
You can write it in Geogebra and you can see the shape of it. It is interesting what does it mean in human's world, economy, information spreading etc
Just asking. Could you use a smaller example of the same concept, and assume the same relationship would apply? For instance could you substitute the question: Which is larger? 4^5 or 5^4? Now you have manageable numbers. 4^5 is larger. So, I would answer that 39^40 is larger, and move on.
you may have a common version, like , which is larger, a^(a+1), or (a+1)^a ?
from video when a>=3, we have (a+1)^a / a^(a+1) < (a+1)^2/ (2* a^2)
suppose (a+1)^2/ (2* a^2) < 1, solve it and get when a>1+root(2)>2, a^(a+1) > (a+1)^a
You could, but in that case you need more than one example to actually find the answer: 2^3 and 3^2 have the opposite case for example, the smaller exponent gives the larger result. A single point isn't necessarily proof yet. Basically you can graph out x^(x+1) vs (x+1)^x, in which x is a real integer, that way you find out that larger powers pick up speed VERY quickly. The difference between 39^40 and 40^39 is more than 4 sextillion, in favor of 39^40.
But yes, it is a solid basis that is way easier to reach the same conclusion with, that might also be more obvious to more people.
Very good solution, the way I did was for transforming 40^39 and 39^40 in (4•10)^39 ○ (3,9•10)^40. Expanding the exponent and dividing by 10^40 in both sides, we get this form: 4^39•10^-1○3,9^40, and 3,9 can also be written as (4-0,1), rewriting it as mentioned before and expanding the exponent: (4^39)/10 ○ 4^40 - 1/10^40, and its clear that the second sentence is bigger than the first one, so: (4^39)/10 < 4^40 - 1/10^40. Let me know if you guys find any error in the solution 😁
That would not work however, since when we expand for example (x+1)^2 we don't get x^2+1^2 but x^2+2x+1^2.
Using binomial theorem you could however expand it and meanwhile prove 39^40>40^39.
This would work like this:
40^39=(39+1)^39 =(binomial theorem)=
39^39 +f(1)*39^38 + f(2)* 39^37 +.... +f(38)*39^1+39^0
Where f(x)= 39!/(x!*(39-x)!) And f(x)
@@isacnordin8048 Thanks for correcting me, appreciate it
👍 to G, your solution in this case is quite clear... 🙂
Can somebody explain the step at 4:28 where we convert all those (1 + 1/39) terms to (1 + 1/2)(1 + 1/3)... etc?
same question here!
That the comparision of (1+1/39) and (1+1/(n+1)) while n=1,2,3... to n=38
He’s not converting them, he’s writing a series that can be compared to determine which series is larger. He selected the series that would let him cancel most of the terms then argued that it was smaller than the original series, therefore simplifying the problem.
that's where the narrator lost me too
@Brad Evans so if I'm understanding right, we want to know if a > b. So we create a new value, c, where it is guaranteed that a > c. And c is also easier to evaluate and prove that c > b. So, since a > c > b, then a > b?
I would just take log for both sides that give 39 log40 O 40 log39 and get rid of the logs by using both sides as the power of 10
Which gives 10^39x40 O 10^40x39
From here the answer is obv but for the record i then take away 10^39 for both sides which will leave us at
40 O 39x10
This doesn't require any calculation. (a+1)^a is always less than a^(a+1). The only exception is 2. 2^3 is less than 3^2.
39^40 = e^(40*ln(39));
40^39 = e^(39*ln(40));
we need to compare right sides of the equations but because they both have the same base e we can compare exponents. So, lets name those new functions:
f(40) = 40*ln(39); f(x) = x*ln(39);
g(40) = 39*ln(40); g(x) = 39*ln(x);
we need to find a point where these functions cross (to be able to count our values from that point)
x*ln(39) = 39*ln(x) - there is an obvious point x = 39;
and now we want to find derivatives of functions f and g
f`(x) = ln(39);
g`(x) = 39/x;
for x ≥ 39 we have
g`(x) ≤ 39/39
g`(x) ≤ 1
f`(x) > g`(x).
it means that f(40) > g(40)
39^40 > 40^39
My job has nothing to do with Math, and have left schooling more than 12 years ago, I just took log of both sides now you have 40log39 and 39log40, LHS>RHS.
This can be done using induction. The general result is that (n+1)^n < n^(n+1)
Нас всегда учили брать ln над показательными функциями: 39/40
39/40
задача заезженная до полного позора. Мне еще в 76-м на вступительном давали сравнить Пи в степени е и е в степени Пи.
Общее решение: (можно логарифмировать, получится примерно то же самое, но менее красиво)
сравним a^b vs b^a где a, b действительные и оба больше или равны е
a^b vs b^a
(a^b)^(1/bc) vs (b^a)^(1/bc)
a^(1/a) vs b^(1/b)
проанализируем функцию y=x^(1/x).
Дифференцируем y(x):
y' = -x^(x-2) * (log(x) - 1), видим, что в е - корень производной, слева функция растет, справа - убывает.
для x > e: y'(x) < 0, следовательно из a > b >= e следует a^(1/a) < b^(1/b), а значит и a^b < b^a
В частности, 40^39 < 39^40, or e^Pi > Pi^e
Еще авторы задач любят подсунуть a=2, а b > e, получается, что числа по разные стороны от максимума - но легко решается с учетом, что 2^(1/2) = 4^(1/4), а значит вместо а=2 можно взять а=4.
Numbers that are (percentage wise) this close together, always go with the larger exponent. Since you're not asked HOW MUCH LARGER, just be happy with going with the larger exponent.
I just did within couple seconds.
2^4 = 16
3^3 = 27
Or maybe better example
2^3 = 8
3^2 = 9
And concluded 39^40 would also be greater than 40^39.
I wonder if there is proof that would support this premise for all real numbers
N^(n+1) < (n+1)^n
I would think the number with 39 as the base is the bigger number because when it's that close, you have to see which exponent is larger. 40 > 39
Yes....logic over calculations✌️
Intuitively, it was easy to know 39^40 is much much greater than 40^39. But what if the question was 40^39 vs 20^40? 🤔
This is direct since 40^39 = 2^39*20^39 and then we know that 2^39 is much bigger than 20 thus 40^39 is much bigger than 20^40. In fact we see at least 40^39 > 20^39 * 20^7 but yeah I get your point. Had the question been about comparing 40^39 and 20^47 then we may might not simplify easily.
@@priyharshgangwar great explanation, thank you
I simply used smaller numbers like 2 and 3 : 2^3 and swapped the numbers.
I get the answer from the 2nd step. Very simple, by a calculator.
Take log with common base on both sides. It will get you which is larger
my intuition only led me to the logarithm world. this method would be awesome!
任何底数为a（a大于等于3），指数为a+1的数一定大于底数为a+1,指数为a的数。例如：5的6次方一定大于6的5次方；18的19次方一定大于19的18次方------，39的40次方一定大于40的39次方，以此类推，无一例外。其实，这可以当作数学的一种公式来使用，不必再做繁琐的演算去证明谁大谁小。如果要证明这一公式的可行性，那就去证明这一公式是否成立。去演算每两个数字的大小大可不必。
if x>e x^(x+1)>(x+1)^x. it is generally solved by differential equation. thus 39^40 > 40^39. no calculation necessary.
What if finding a pattern like 3^4>4^3... 4^5>5^4 and so on 39^40 must be bigger
Démonstration by récurrence can work as well, N^(N-1) is smaller than (N-1)^N for all N>4
It's trust for othwr numbers as well. It trust for 2, 3, 4, 5 etc
@@hariselezovic6-4 hu, no. 2 > 1, 9 > 8 but 64 < 81 .
السلام عليكم من المغرب
شكرا على مجهوداتكم
حاولوا التنويع من مقارنة وحل معادلة وتحديد قيمة قصوى ............
If you apply log both sides, it's pretty easy to find out.
The 39 member product approximates e which at around 2.73 something is much smaller than 39
這個方法真的漂亮，用log解要做好麻煩的呢
just use binomial approximation once u reach (1+1/39)^39
To calculate \((1 + \sqrt{5})/2)^{12}\), we can use the binomial theorem or simply expand it out.
Using the binomial theorem:
\[(1 + \sqrt{5})/2)^{12} = \binom{12}{0}\left(\frac{1}{2}
ight)^{12} (\sqrt{5})^0 + \binom{12}{1}\left(\frac{1}{2}
ight)^{11} (\sqrt{5})^1 + \binom{12}{2}\left(\frac{1}{2}
ight)^{10} (\sqrt{5})^2 + \ldots\]
We can simplify this:
\[(1 + \sqrt{5})/2)^{12} = \left(\frac{1}{2}
ight)^{12} + 12 \left(\frac{1}{2}
ight)^{11} \sqrt{5} + \ldots\]
However, calculating all the terms might be cumbersome. Let's use a calculator to simplify:
\((1 + \sqrt{5})/2 \approx 1.61803398875/2 \approx 0.80901699438\)
Now, raising this to the power of 12:
\((0.80901699438)^{12} \approx 0.07957747154\)
So, \((1 + \sqrt{5})/2)^{12} \approx 0.07957747154\).
Im glad ive saved the cost of your services
To use the inequality at 7:12, you need to ALREADY know the correct answer. That is, if the correct answer turned out to be the other direction, this inequality at 7:12 would do you no good. I think that the suggested methods using binomial approximation or the definition of e are simpler and better.
With two numbers that close, of course the larger power will give the largest number.
That is only partly true because 3^2 > 2^3 i.e 9>8 so bigger base can give a bigger result too
but 3^2 > 2^3
@@jedrzejwrotynski9421 Only when you are close to the origin on the number line.
@@jeffrybassett7374 Not only. Bigger negative bases in smaller odd negative powers give bigger results, too. For example (-10)>(-100) and (-3)(-100)^(-1) i.e (-1/1000)>(-1/100). And -10 or -100 are not from the begining of the number line! qed
4:25 - Could someone please explain to me how you're allowed to just change (1 + 1/39) into (1 + 1/2) ... then (1 + 1/3) ... etc? This step makes no sense to me and is not explained.
since it is always smaller than (1+1/2)(1+1/3)(1+1/4)...(1+1/39) and therefore if I get this, many numbers which appear on the denominator and on the numerator as well can cancel out and it will be easier.
pretty intuitive but great proof!
if a>e then a^(a+1) > (a+1)^a it is easy to proove with calculus if you using f(x)=(lnx)/x .... so if a=39 ......>>>>39^40> 40^39
Used logarithms is easier.
why lot compare the numbers logaritms instead and compare those?
ln(40)^39=39ln(40) and ln(39)^40=40ln(39)
Now its easy, ln(40) is just tiiiny bigger much less than 1 than ln(39), so now clearly since 40-39=1 we can say that 40ln(39) is bigger that 39ln(40). 39^40>40^39.
Hey ya, i think in divides 39^40 : 40 ^39 = 39 * ((39/40)^39) = 39 * ((1 - 1/40)^39), then use inequalities look like yours
once you get to the third step (in red) above, it is simple by induction to take (40/39)^n and find where N>1. N is way more than 39. You dont have to waste all this time on the proof to know the answer.
Я смотрел много таких при метров, и всегда больше число с большим показателем. Я думаю, это можно доказать как теорему и перестать отнимать наше время и, соответственно, деньги.
40^39 is 302 novemdecillion while 39^40 is 4.39 vigintillion so 39^40 is bigger
Why is this so hard to solve if you can just put 39^40=39•39^39 which clearly has a bigger base than 40^39 meaning it has to be larger? Why is there such a hassle also I don’t understand where the whole blue part came out of…
Why would you do this to yourself?
You made your solution more complicated & hard to follow.
The exponent is more powerful than the base, so it's evident without calculation.
Спасибо.
thanks!
Too complicated demo
Lim (1+1/n)^n =e
n->♾
(1+1/39)^39 e/39
( 1+1/39)^39 will be less than 2.71 rather "e" so the answer is just one step e < 39
The no. Whose power is bigger is always the answer
If a > b, then b^a > a^b
As a and b > e
Never forgetti💅
Great !
Without watching the video nor reading the comments I'd try to prove a general theorem by a don't-know-the-result kind of induction: Assume (n+1)^n : n^(n+1) holds, let's prove (n+2)^(n+1) : (n+1)^(n+2) (where : is either < or >). To get from a=(n+1)^n to b=(n+2)^(n+1) we multiply by b/a, hence we have
(n+2)^(n+1)
: n^(n+1) * (n+2)^(n+1) / (n+1)^n
= ( n*(n+2)/(n+1) )^(n+1) * (n+1)
= ( n*(n+2)/(n+1) )^(n+2) * (n+1)² / n(n+2)
= (n+1)^(n+2) * ( n*(n+2)/(n+1)² )^(n+2) * (n+1)² / n(n+2)
= b * c
: (n+1)^(n+2)
To find out :, we have to find out if c is bigger or smaller than 1:
( n*(n+2)/(n+1)² )^(n+2) * (n+1)² / n(n+2)
= ( (n²+2n) / (n² + 2n + 2) )^(n+2) * ((n² + 2n + 2) / (n² + 4n))
Consider big numbers. Then (n²+2n) / (n² + 2n + 2) < 1. Since we exponentiate, this term dominates our calculation, hence : is
I love the solution of the other comments regarding using e. I didn't even think of that, although I know it. Also induction was probably a very bad choice, but anyway, I got a right result.
Nice job mate. I have a habit of using logarithm to I didn't even think of this kind of approach
@@redeyexxx1841 I'm not proud of it tbh, many other approaches mentioned in the comments are way better.
Bruh, you could have ended the problem at taking the powers common. It would be something like (1.x)^39 *1/39 , which is very much less than 1
If you need answer only which one is larger, then…..
3^4 = 81. Larger than 4^3 = 64
So 39^40 must be the larger one
This is a very common type of question on channels like this. So I plot x^(x-1)-(x-1)^x. It is negative for all x > 3.293. Can we get on with some interesting math problems now finally?
With z=x-1 then it seems that x^(z/a) - (z/a)^x = 0 when x = 3.293 for any a. Is 3.293 (rounded) a special number?
It's the number where (x-1)^x = x^(x-1), and we call it the solution of the equation...
Probably as special as the real number of x that satisfies e^(-x) = 2x.
I just think "10^2 vs 2^10" and it becomes obvious.
i believe in the basics of mathematical statistic, exponent always has bigger growth value, than any operand. no need of any calculation
So simple 39^40. Why do you need 10 mn and 2 s for that ?
40 with 39 zeroes at the end will be smaller than 39 with 40 zeroes at the end.
😆😆😆
a^b always smaller than b^a, if a>b?
You can also sove like this To compare \(40^{39}\) and \(39^{49}\), we can take the natural logarithm of both and compare their logarithms.
Let's denote \(a = 40^{39}\) and \(b = 39^{49}\).
\(\log(a) = \log(40^{39}) = 39 \cdot \log(40)\)
\(\log(b) = \log(39^{49}) = 49 \cdot \log(39)\)
Now, we can compare the values of \(39 \cdot \log(40)\) and \(49 \cdot \log(39)\):
\[39 \cdot \log(40) \stackrel{?}{
Power means multiple.