Comparing: 100^99 and 99^100, which is larger?

How to solve this maths question in the best way? Watch the video and find out the method!

[@PXO005]

For any 2 numbers greater than e, the one closest to e will be the greatest when raised to the other number's power

[@davidseed2939]

yes and the reason is that comparisons of this type can be transformed into comparing x^(1/x) with y^(1/y) in this case 100^(1/100) cf 99^(1/99). 99 is the smaller number and thus 99^(1/99) is the greater

[@CAROLUSPRIMA]

I have an earned doctorate and you guys are impressing the hell out of me. I got through graduate level statistics and still don’t know how.

[@kapsel-yg2sk]

Yes, and I'm disappointed that youtube is still flooded with this class of problems

[@FlummoxTheMagnificent]

So basically the smallest to the power of the largest, if they are greater than e.

[@rachityadav5011]

@@CAROLUSPRIMA we will get this by the graph of F(x)=lnx/x and we will get that func is increasing upto e and decreasing after e then just putt values then just cross multiply and take numerical part in the power of log and than cancel out log you will get your answer

[@cameronspalding9792]

Take derivatives of ln(x)/x to find that the function is maximised when x=e. ln(99)/99>ln(100)/100 so 99^100 is bigger

[@ToiLaChinhMinh]

ye 99^100 = 99 multiply by itself 100 times. And with 100^99 that's just 99 times so😅

[@kkkkjlkkkkj]

bro why tf you just do 100^99(99+1)^99 99^99+1 that is visually smaller than 99^100

[@Jinsun202]

@@kkkkjlkkkkj "visually" 🤣🤣🤣

[@kakeruhirota-m5d]

This solution method is popular in Japanese entrance examination.

[@abrahimbacil2454]

You have to learn to write 9 instead of 8

[@vjlaxmanan6965]

Yes...agree! :(

[@yeshwantphatak4709]

😀😀

[@bobbylashley5995]

It's understandable so stop complaining

[@tomlee9534]

9 instead of g I reckon.

[@chillingsk]

Nah that's the gravitational constant

[@rithvikmuthyalapati9754]

If e

[@davidseed2939]

the best approach is to divide each side by 99^99. then R=99 L=(100/99)^99 =(1+1/n)^n for n=99 but.from the definition of e L

[@keinKlarname]

You need that (1 + 1/n)^n is increasing when n is increasing.

[@davidseed2939]

@@keinKlarname which is true. but i only need that the sequence is bounded above which is also true.

[@ericwilliams1832]

awesome!

[@brian554xx]

i think my brain is becoming inflexible. i had to keep reminding myself that the characters that looked like 'g' are actually '9' and the characters that look like 'Λ' are actually '1'. when I was younger I would have seen the pattern and just kind of flipped a switch in my brain to be in your local writing style. I'm getting old. i hate it, but it beats the alternative.

[@kenhaley4]

It is sometimes a problem as her 9's sometimes look like 8's, as the tail of the 9 comes up and touches the upper loop.

[@photondance]

It’s very simple. 100^99 = error 99^100 = error 100^99 = 99^100

[@shinybagel3407]

Lol

[@Thorfinnkarlsefni047]

Omg it's Albert Einstein

[@deleted-something]

Genius

[@thonydq]

This is elon musk

[@adhamabdelaty2479]

bruh

[@cameronspalding9792]

One can prove (1+1/n)^n

[@marklevin3236]

Whenever e100^99. Proof is not hard Let's f(x)=ln x / x... For x >e this function is decreasing since derivative is negative. Therefore ln a /a > ln b/ b. If we multiply this by ab в lna > a ln b. ln (a^b)> ln(b^a) a^b > b^ a

[@marcusdecarvalho1354]

#thanks

[@aca4262]

So you are saying 5^2 is bigger than 2^5?

[@aca4262]

Is that it? cause i'm confused

[@marklevin3236]

@@aca4262 a and b must be greater than e. 2

[@aca4262]

@@marklevin3236 Tnx got it.

[@elenabale5755]

Once at a school math olympiad, I got a task to compare 2007^ 2008 and 2008^2007. 15 years have passed, and I still don't know how to solve it

[@vigneshwars1331]

@Dylan Bradley that would be the right answer because multiplying 2k something extra will change the number largely

[@elenabale5755]

by the way, now I have known how to solve it 🙂 youtube really helps sometimes

[@locomotivetrainstation6053]

2007^2008 is biggest

[@qtfy]

@Dylan Bradley ok but why that didn't work on the video tho

[@Jinsun202]

For any two consecutive numbers, the expression with the smaller base will be greater for bases ≥ 1+√2. For bases less than 1+√2 the greater based expression will be greater.

[@MikkoRantalainen]

If you look at a still frame around 2:50, you'll see a nice demonstration why you shouldn't use a hook in number 9 because it's really easy to mix it with 8. I've intentionally trained myself to avoid the hook in the bottom of the 9 to make it harder to mix it with 8.

[@momonga2850]

Man...she said that it was 99..... Please stop complaining and atleast thank her that you came to know something interesting

[@alessandroonorato4966]

I think there's an easier way to solve it: let's consider b = a - 1. 1) When a=4 and b=3, a^b=64 and b^a=81. 2) When a=5 and b=4, a^b=625 and b^a=1024. Then 1) b^a - a^b = 81 - 64 =17 2) b^a - a^b = 1024 - 625 = 399 This means that b^a grows faster than a^b. So considering a=100 and b=99, 99^100 must be bigger than 100^99

[@arion6649]

not enough

[@gytisdramblewolfskis8521]

I compared 3^4 vs 4^3 and guessed that anything higher would have the same rule

[@newthan2]

same

[@Jinsun202]

"guessed" 🤣🤣🤣

[@shrikrishnahospital9762]

Very good

[@npcwithsidequest]

Yeah, that is pretty much it actually.

[@homesynthesis]

the general solution is if e

[@admink8662]

I think you need to proof there exists N < 100 such that (1+1/n)^n increasing for n > N

[@nickcampbell3812]

Reminds me of a blackpenredpen video. Main takeaway is if a

[@anonymouseniller6688]

I like the way you explained it, even more than the problem itself. Thanks for sharing, I shall look up more of your videos.

[@northdallashs1]

Clever, but you’ve left out two things. 1) How do you know that (1+1/n)^n is an increasing function of n? 2) How do you know that your limit = e? Or for that matter, how do you know that (1+1/n)^n is even bounded as n approaches infinity.

[@haotiankong6904]

I think she thought both of them were given. Most people who know what limits are probably already know about this specific limit.

[@joseluisvazquez4221]

The limit can be a given. But the increasing function fact is not at all

[@EatThatLogic]

This is a good solution. I had recently made a video on e^pi vs pi^e with a different approach.

[@Mr.kasugai]

oh,that's famous problem in Japan because this problem afflicted examinee on entrance examination of elite university in Japan.

[@EatThatLogic]

@@Mr.kasugai Now you know where the solution is :)

[@WealthWonders777]

You can just 100⁹⁹ vs 99¹⁰⁰ 10¹⁰⁰ vs 99¹⁰⁰ 99 is bigger

[@rls5907]

@ WealthWonders777 Your maths is incorrect. 100^99 = (10^2)^99 = 10^101. It’s wrong to write that 100^99 = 10^100

[@binhanh296]

In a math problem where we have to compare the value of x^y with y^x where y=x+1, let's say that x^y=a and y^x=b, only the pair 2^3=8 < 3^2=9 where we have a

[@tarikyalcin2819]

İ agree with you. Clever way.

[@LD-dt1sk]

Me with a calculator: I am 4 parallel universes ahead of you

[@shreyjain3197]

pretty sure no calculator can compute 99^100 but ok

[@LD-dt1sk]

@@shreyjain3197 you can just use Google.

[@calvinsebastian8902]

2³ = 8 3² = 9 But the next sequences is 3⁴ = 81 4³ = 64 So i think 100^99 < 99^100

[@gameflame4633]

Sorry if there are any mistakes here, english is not my native language. I made a ratio of 99¹⁰⁰/100⁹⁹, got 0.99⁹⁹×99, put 0.99 as an approximate value of 1 and from there I saw that the ratio was greater than 1 and concluded that 99¹⁰⁰ is greater than 100⁹⁹. Edit: I allow you not to write the same type of comments about not rounding 0.99 to 1.

[@a_minor]

your english is still better than most native speakers

[@cringeconnoisseur6037]

@@a_minor he missed a comma 🤬

[@shaguna]

When you approximated up from 0.99 to 1 you introduced a margin of error, and you cannot use that approximation to just state that the initial number will be higher. For example, consider the function f(x)=x*[(0.9)^x] (x multiplied by 0.9 to the power of x). f(33)=1.018804.... and f(34)=0.945636 .... According to your method, if we round up, both values should be bigger than 1, but they are not.

[@cringeconnoisseur6037]

@@shaguna if he approximated .9 to 1 he would have been completely wrong as well.

[@vadirajpogal4199]

since 1*1=1 any number less than 1 multiplied by itself, product should be less than 1. so 0.99 raised to 0.99 has to be less than 1 is a logical conclusion and not an assumption is my humble opinion 🙏

[@pinedelgado4743]

Amazing!!! Whoever you are, I wish I had YOU for a math(s) teacher when I was in my secondary and post-secondary school years!! BTW, I've just now liked this video, subscribed to your channel and requested ALL notifications to it!! Thank you very much!!

[@shreyjain3197]

As a general rule, a^(a+1) is always greater than (a+1)^a

[@trigeminalneuralgia9889]

not really, it's opposite for values of a less than e, 1^2

[@shreyjain3197]

@@trigeminalneuralgia9889 ok fine but note that I did say "general" and also if it's such small values just calculate it yourself

[@jasonrichner2537]

You're both right. It works for 3 and above.

[@pablojp3498]

@@shreyjain3197 rather than say in general say. When a>2.29317, a^(a+1)>(a+1)^a

[@Jinsun202]

@@shreyjain3197 It's certainly not general since there are infinite values less than e.

[@prashanthacharya5753]

For any 2 numbers x, y greater than e where x > y … y ^x > x^ y. . One can prove this by differentiation

[@henrytang2203]

In my mind, it's: 99ln(100) vs. 100ln(99) And ln() grows slower than linear for larger numbers. So 99^100 is bigger.

[@karlisberzins5676]

you can use same approach as in the 1.005 vs 2 video -> first take the 100^99 convert it to (99*(100/99))^99 which then equals 99^99 * (1+ 1/99)^99 and then compare the (1+ 1/99)^99 to 99. then you can expand the 99 to: 2/1 * 3/2 * 4/3 ... 98/97 * 99/98 to get the 99th fraction we can split 2/1 into sqrt of 2 which is around 1.41 -> we get sqrt(2) * sqrt(2) * (1 + 1/2) * (1 + 1/3) * (1 + 1/4) ... (1 + 1/97) * (1+ 1/98) and then compare these fractions to the (1+ 1/99)^99 and we can see that every fraction in the first set is larger than any in the second one therefore 99^100 is larger. there is probably better way to get 99 fractions, and splitting the "2/1" seems crude, but it works as 100/99 is much less than sqrt(2)

[@elir626]

Rather than using complex math, just use logic, if 2^3 is bigger than 3^2 then the same logic must apply to every other scenario accept for 1^2 vs 2^1

[@karlisberzins5676]

@@elir626 2^3=8 < 3^2=9 it seems that after around 2.29^3.29 vs 3.29^2.29 you are right, but that would have to be shown (i just plotted the y=x^(x+1)-(x+1)^x in wolfram alpha... modern tools are cool)

[@elir626]

@Kārlis Bērziņš oh shoott I think I was thinking of 3^3 as 27 and not 2^3

[@nirvairsingh3416]

Meanwhile me solving it using binomial theorem

[@JPTaquari]

Easy, it's not to compare values, but simply to see which is the biggest, so... 100/99 = 1,0101 99 / 99 = 1 100¹ = 100 99^1,0101 = 103,7 Bingo !!!!!!!!!!!!!!!

[@tamimplayz]

her way of writing "9" and "lim" concerns me

[@lunarthicclipse8219]

We could just change the 100 ^ 99 into a 99 ^ Log99(100) * 99, since the value inside the log is greater than the base, so the value of Log99(100) must tend to 1 from the right side as it reaches 99, and since 100 is just 1 more than 99, that means that its really close to 1, so we can assume that log99(100) is ~1, so by substituting 1 for log99(100), we get 99 ^ 99 * 1 which is smaller than 99^100

[@Shekhuguruji]

absolutely amazing mind, you've a great engineered mind, thank you for this

[@lunarthicclipse8219]

@@Shekhuguruji oh thank u!

[@vikrantbajage]

Well, you can apply the logic as 100 = 2 x 2 x 5 x 5 and 99 = 3 x 3 x 11 so addition of their factors would be for 100 ~ 14 and 99 ~ 17 so if the powers were to consider 14² will be always less than 17³. So 100⁹⁹ will always be less than 99¹⁰⁰.

[@lynxprime1778]

For the people who dont want to see the whole video, the answer is 99 to the power 100

[@thirunavukkarasuasaimuthu6799]

Madam , just take log of both and convert to similar exponent for both terms, it quick to see second term is bigger with anti log . Of both terms, we just need log of 2 and simple log and antilog

[@toco1318]

Exactly what I was thinking

[@bryanekers3472]

@@toco1318 That was my approach, also. 100 = 2^6.6439, therefore 100^99 = 2^(6.6439 x 99) = 2^657.742. 99 = 2^6.6294, therefore 99^100 = 2^(6.6294 x 100) = 2^662.94 Therefore 99^100 is larger, and we can even easily tell by how much, i.e. it's 2^5.198 times larger, or 36.707 times larger.

[@akatsukiesports-aks-7859]

It's simple if you make both numbers to the same exponent, the number with the largest base is larger.... Therefore (99x99)^99 is larger than 100^99

[@QuartzQuill]

99^100 is bigger though... They don't have the same exponent, that's the whole point of the question...

[@akatsukiesports-aks-7859]

@@QuartzQuill 99^100= 99x99x99.....100 times. Or if you multiply once and make the rest the exponent, you get (99x99)^100-1 Which simplifies to 9801^99

[@QuartzQuill]

@@akatsukiesports-aks-7859 just do 99^100-100^99, you get a positive number therefore 99^100 is bigger

[@akatsukiesports-aks-7859]

@@QuartzQuill no

[@akatsukiesports-aks-7859]

U using a calculator for this?

[@kimamaral3015]

another way to solve this sort of problemn without calculator is using the expansion of (a-b)^n which is commonly known by people who have gone through calculus 1 classes (as to prove the derivative of x^n). This way we can turn 99^100 = (100-1)^100 = 99 * (100^99 + 100^98 + ... + 100 + 1) [using the formula of the expansion of (a-b)^n], and this is obviously bigger than 100^99 since this exact term appears in the sum of strictively positive terms and is multiplied by a number bigger than 1 (99), so with that it gets crystal clear that 99^100 >> 100^99.

[@parthdurgude2617]

We could simply divide both nos. by 99⁹⁹ so 100⁹⁹ would become (100/99)⁹⁹ which would be around (1.01)⁹⁹ which will give 2.67 (

[@hectorurdiales4570]

100*log99 > 99*log100 Since log is a strictly increasing function in the positive numbers (and 99 and 100 are both positive numbers), then 99^100 > 100^99

[@Pao-vo8mf]

listen, nerd, your parlance is uninteligible. harness the power of concrete words!

[@robertveith6383]

You did not justify that top inequality. log(100) > log(99), so you would have to explain how the right-hand side is not large enough to be larger than the left-hand side.

[@Lowraith]

2^3 vs 3^2 ...

[@Piotr_Tokarz]

I used binomial theorem and arithmetic mean and got the same result. I feel happy. Also I discovered a type of formula for these types of inequalities.

[@gddfx8513]

Super piotruś

[@BRUH-sk4lp]

Can we do this using binomial theorem?

[@ligion324]

This kind of question can be generalized and all have a common answer, (n+1)^n is always smaller than n^(n+1) when n>2 (n is integer)

[@bappabain2142]

General method (one way-fit all) is to take log log(100^99)=99log100=198 log(99^100)=100log99=199.5635 so 99^100 is larger. (simplest method)

[@wiggles7976]

Then you are relying on whether or not a calculator is accurate enough, and this assumption can fail in other cases. Getting a proof is the only way to be sure.

[@bappabain2142]

@@wiggles7976 2 decimal place of log 99 is enough to show the difference....high accuracy not reqd....though all calculators are accurate now upto atleast 10 decimal places

[@wiggles7976]

@@bappabain2142 For this case it works, sure, but the point of the question is to be able to do it without the help of a calculator.

[@bappabain2142]

@@wiggles7976 yes for that we need classical maths

[@naniwas_tiger]

idk what the video is all about but can't we just solve it by using 100⁹⁹×100/99¹⁰⁰×100=100¹⁰⁰/99¹⁰⁰×100 and using law of exponents (100/99)^100×1/100. 100/99 is 1.01 so 1.01¹⁰⁰/100 and since 1.01¹⁰⁰/100100⁹⁹

[@TheVirtualArena24]

I'm very poor with math. Why do we divided both in first place if we wanted to see which one is bigger? And at last part how 1+1/99(99) is

[@midievel8479]

Easiest question ever. Exponents are so much bigger than the difference between 99 and 100. So it should be clear that 99 getting an extra ability to double itself would be greater than 100 ^99

[@gytisdramblewolfskis8521]

Good point.

[@anonymousperson4466]

even without the approximation it is clear that (1+1/99)^99 will have expansion in the form 1+(99)(1/99) + (99)(98)(1/99)²+.... the rest of the terms will be too small to make difference....and the above number is just around 2-3 so fraction will be 2.../99

[@elir626]

And I knew without doing any complex math bc if 3^4 is bigger than 4^3 than that should apply all the way up to thos scenario

[@anshumanrath6703]

When bases are similar, larger power is always bigger especially in large numbers

[@nguyentrongnhan6908]

Actually, with x larger than 2, x^(x+1) is always larger than (x+1)^x

[@adamslajer2203]

@@nguyentrongnhan6908 actually not larger than 2 but larger than e(2.718...)

[@dokangmath]

sequence a_n=(1+1/n)^n < 3 bdd and finish with n=99

[@Lowraith]

x^n > (x+1)^(n-1) for all values x > 2, and n >= 1

[@trevorwilliamsmith]

another way: 100^99 or 99^100 ; divide by 99^99 (100/99)^99 or 99^1 (1+1/99)^99 or 99 from the 1st order binomial approximation (1+x)^n ≈ 1+nx 1+99/99 or 99 2 or 99

[@st.charlesstreet9876]

Thank you for the bringing back the logic! Love this post🎉

[@iamblablabla]

It can be solved by logarithmic numbering Of course 100 pow 99 greater than 99 pow 100 100 pow 99 is 99 log 100 99 pow 100 is 99 log 99 + 1 The ratio is 99 log 100/99 + log 1/99 more than 1

[@Pro_Tro]

I coded a program to print the total amount of digits in 100^99 and 99^100 And I found that 100^99 has 199 digits But 99^100 has 200 digits so 99^100 isbgreater

[@smaug9833]

Just write it as (1+1/99)^99 x 1/99 = 2/99 < 1 (1+x)^n = 1+nx for large values of n and x

[@colinmccarthy7921]

I looked up the answer on the Internet. If I am correct,it said that 99(100) is greater than 100(99).

[@mrtrinity143]

I did also my own computation and the result was 99^100 was the greater..🙄

[@dawgslice5745]

I understand everything except for why the 9s look like either an 8 or a g

[@pranaykumar5683]

Way of teaching is also very excellent

[@tamirerez2547]

GENERAL SOLUTION: For any M and N (None of them= 1 or 2) If M > N then N^M > M^N ALWAYS. (The expression with the SMALER base is BIGGER) Therefore 99^100 < 100^99

[@drozfarnyline4940]

Exactly i was thinking about this general case which killed it in seconds

[@haotiankong6904]

I'm pretty sure it works the other way around when it's smaller than e. I guess if your criteria is that both M and N have to be whole numbers, then I guess it works. I think generally, the base that is closer to e is bigger.

[@oniondesu9633]

what was the point of writing e to 17 significant figures and explaining how easy it is to remember them? all than matters in this problem is that e is less than 99

[@oniondesu9633]

also, it really should be shown in this solution that (1+1/n)^n is a strictly increasing sequence, even if it does seem trivial

[@robertgumpi7235]

It’s easier to divide the inequation by 99^99. Then you have e to the left and 99 to the right. So left is smaller.

[@robertveith6383]

You won't have e on the left-hand side, but you will have under approximation to it.

[@lamchekyeow]

Well, it is quite obvious 99^100 is bigger because of the yield of e > 1 I have no idea why there is a need to have so much additional equations, tho

[@sahaajsinghchawla1307]

I thought about making the 99^100 as (100-1)^100. Which then becomes 100^100+1^100-(2×100×1). Which makes it 100^100-199 and this is bigger than 100^99

[@adityasingh3963]

That indentity that you are using to open the brackets only works when you are squaring. For example (a-b)² = a²+b²-2ab, but (a-b)³ = a³ - b³ -3a²b + 3ab². Similarly (a-b)¹⁰⁰ has a different formula. It's not as simple as you thought. Though you got the answer correct the method you used is wrong. And it can give completely wrong answers for comparing numbers like 100⁴⁹⁹ and 99⁵⁰⁰. According to your method 99⁵⁰⁰ should be larger but in actuality 100⁴⁹⁹ is larger.

[@kimamaral3015]

@@adityasingh3963 if he used the identity of (a-b)^n it would be correct, that is how i've done it. Im just not sure if this is a high school problemn, because if it is, im not sure if in high school this identity is as widely known as in calculus

[@yet4600]

Good development. But as a contribution I must comment that, in this case and in other similar cases, the largest is always the one with the smallest base and the greatest exponent.

[@kingalexisdumaquita8284]

math got confusing when she said -the number 'e'-

[@omchavan5664]

Well maybe i am wrong, but if you open a calculator and try something like 3^4 and 4^3,. 3^4 is greater again try 5^4&4^5, 4^5 is greater and how many ever you try for numbers above 3 you will always get the smaller number with greater value , so 99^100 should be greater Ofc numbers 1&2 and 2&3 are exceptions

[@DB-lg5sq]

شكرا لكم على المجهودات دون استعمال e نستعمل 100^99/99^100 =(100/99)^100× 1/100 =(1+ 1/99)^100× 1/100

[@danielstatler954]

99^100 is just (100*0.99)^100. Take out 100^99 and cancel out the 100^99 on the top. Leaving us with 1 / 100*0.99^100. 100*99^100 is probably bigger than 1.

[@vikramk6869]

a^b > b^a For e

[@spectorworld]

Can we look at this in a different way? 100^99 = (99 + 1)^99 = 99^99 + 1^99 99^100 = 99^99 × 99 By replacing 99^99 with a constant a, it becomes obvious that: a + 1^99 is much smaller than a × 99, Right?

[@sniggleboots]

(99 + 1)^99 is not equal to 99^99 + 1^99, you have to use the binomial expansion formula

[@marcusbighouse]

I watched it without sound and took me a while to understand that 89 is supposed to be 99

[@otakurocklee]

You need to prove that (1+1/x)^x is always increasing for n>=1

[@abhinayudnur3319]

It's very simple. When both the base number and exponent are equal, the number having the highest exponent will yield the highest result.

[@konchady1]

There's a more useful fact you did use but didn't correctly prove. That is, this sequence is monotone increasing. The monotone increasing part follows from a clever application of AM-GM inequality. Thus e is not only the limit but also a valid upper bound. In some books, it is easier to show the sequence is upper bounded by 3. That also works.

[@laizerwoolf]

I tried it with smaller number like 9^8 vs 8^9 and calculating it manually, to arrive at a similar answer.

[@Lucifez6160]

Yes, when a number is positive and higher than e (2.71828), the number with the higher exponent will always be higher. But if it is below the e, the higher base will always be higher.

[@mihaipuiu6231]

Doamna Math Window...the proof it liked me very much, but be more careful when you write digit 9, which in a few locations I confuse 9 w/ 8. So, please be a little bit careful. And good luck!

[@mathwindow]

Thanks for your remind!

[@ESU01USER]

I thought 100^100 is bigger than (99^100)x100, but it was wrong :-o

[@ipaki6576]

I am still finding which is greater.

[@jcawly1]

If this is a simple question of determining which is larger, then the really easy solution is to convert both to scientific notation with the same exponent. 100E99= 1.0x10^101, and 99E100=9.9 x 10^101. So, it is easy to see that 9.9 x 10^101 > 1.0 x 10^101.

[@cutewhitepaws]

Wao Genius ... But 9.9 x 10^101= 99 * 10^100 which is not same as 99^100 ,

[@Rhys_1000]

100⁹⁹ = 198 zeros or 199 digits 99¹⁰⁰ = 3.66032341e199 which is 199 digits so 99¹⁰⁰ is larger than 100⁹⁹

[@great5832]

Bro.... u literally answered it in the simplest way 👍

[@Rhys_1000]

@@great5832 Im like Khaby Lame

[@DilawarKhan-xv5ye]

100⁹⁹ and 99¹⁰⁰ Take log of both, log (100⁹⁹) and log (99¹⁰⁰) 99 log 100 and 100 log 99 99 × 2 and 100 × 1.9956 198 and 199.56 198 < 199.56 Since equality doesn't change while taking log, so 100⁹⁹ < 99¹⁰⁰

[@aashaylad4545]

Simply apply log on both sides

[@michaelwarren5084]

Why you write g's for 9's ?

[@lorenzed1]

It's simple 100^99 = 10^100 10^100 < 99^100

[@downburst3236]

Says 99, writes 89. But has no problem with 1s and 0s. Looking forward to binary problems.

[@michaelkang215]

You had me at 2.718 😂. - from Jerry Maguire.

[@SassePhoto]

Chat OpenAI: To determine which of these two numbers is larger, we can use the property of exponents that states that for any base x, x^a > x^b if a > b. In this case, 99 > 100, so 99^100 > 100^99. Therefore, 99^100 is larger.

[@mikatu]

your math is awful, since when 99 > 100 ??

[@MinhTran-ri8mr]

You have to prove that sequence (1+1/n)^n is increase

[@1nguoixauxi2]

well 😂 my solution is compare 2^3 and 3^2 😂 the solution for the question should be the same

[@ybrbnf333]

how can we prove that (1 + 1/99)⁹⁹ < e?

[@Evgeny-Kasintsev]

Solution for physicist not for mathematician

[@bartsimpson81]

Very good solving

[@tachles_math]

Lol what a savage for remembering so many numbers after the dec point of e

[@Yougottacryforthis]

100%

[@mir-11]

We can just log both sides: 99*log100 vs 100*log99 198 < 199..

[@RobertShane]

The calculator on Windows can handle numbers this large so I used that

[@darknight3613]

You can just try to bring both sides to have the same power. 99^100 = (99^(100/99))^99 = 103^99 > 100^99 and thats it

[@dougnettleton5326]

How did you get 99^(100/99) = 103, without a calculator? And by the way, it's not.

[@darknight3613]

@@dougnettleton5326 i did use a calculator, its 103.7 or something like that, which is greater than 100

[@chandrakanthsangewar9341]

If (1+1/n)*n = e, then why is (1+1/99)*99 smaller than e when n can be any number?