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Пікірлер
@smaug9833 9 сағат бұрын
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<<1
@anthonyvalenti9093 21 сағат бұрын
I graphed y=3^x and y=x^3 by hand. They are equal at x=3. And y=3^x is greater after that.
@CalculusIsFun1 Күн бұрын
Took me about 2 seconds. Not hard at all in my opinion.
@user-ox9yb1qr6l Күн бұрын
Я воспользовался дифференциалом функций √64+1 и √64-1. {1/(2√64)}-{-1/(2√64)}=(1/16 )+(1/16)=0.125 с точностью до 0.000001😊
@Kriorem Күн бұрын
3 is less than pie...
@davidmilhouscarter8198 Күн бұрын
I haven’t watched the video yet. My guess is 3 raised to pi is larger because the exponent is larger.
@costarich8029 2 күн бұрын
Really simple rule. If a and b are both greater than e (2.71828...) then the expression with the bigger exponent always wins.
@jejojoje9521 Күн бұрын
And otherwise, the bigger base always wins?
@costarich8029 Күн бұрын
@@jejojoje9521 I tried to include a link, but it seems to have gotten clobbered. If both terms are less than 'e' then yes you are correct. Largest base wins. Otherwise it is complicated as the crossover point follows a fussy curve involving the Lambert W (product log) function. I'll try to post the link to the graph in the next comment to see if it's just my inclusion of the link causing it to get deleted.
@costarich8029 Күн бұрын
Yeah the comment with the link gets auto-deleted. The critical curve where x^y=y^x is solved by using the Lambert W function. x = -yW(-log(y)/y)/log(y) which gives a curve that looks like it asymptotically approaches (1,∞) and (∞,1) while going through the point (e,e).
@rolandpokorny2813 2 күн бұрын
3^3,14 is greater
@Leonhard-Euler 2 күн бұрын
pi^3 = 3^3*(1 + (pi-3)/3)^3 = 3^3 * (1 + (pi-3)/3)^(3/(pi - 3) * (pi - 3)/3 * 3) = 3^3 * ((1 + (pi - 3)/3)^(3/(pi - 3))^(pi - 3) < 3^3 * e^(pi - 3) < 3^3 * 3^(pi - 3) = 3^pi
@xanderlopez3458 2 күн бұрын
Thank you for the consistent posts. I get a little smarter every time. Your explanations always seem to come together, even if it takes me half an hour or so to get some concepts. 😊
@davidellis1079 2 күн бұрын
Your solution is more elegant than mine, but here is the way I solved it :- a^5 + a^4 + 1 has no rational roots, so rather than looking for linear factors try a factorisation of the form (a^3 + ba^2 + ca + 1)(a^2 - ca +1) . Then the coeff. of a is c - c = 0 ; the coeff. of a^2 is b - c^2 + 1 = 0 ; the coeff, of a^3 is 1 - bc + c = 0 ; and the coeff, of a^4 is -c + b = 1 . Solving for b and c gives b = 0 and c = -1 . Hence the required factorisation is (a^3 - a + 1)(a^2 + a +1) .
@kevindegryse9750 3 күн бұрын
Great. I would have appreciate to have the justification of why the greater than symbol does not flip at any point (common mistake with inequalities) (multiplication by positive numbers, ln function continuously increasing). But really great overall.
@kb27787 3 күн бұрын
ummmm x = 1?
@davidseed2939 3 күн бұрын
Take powers of 1/(3π) 3^(1/3) ~ π^(1/π) if cam be shown that the function x^(1/x) has a maximum at x= e and is monotonic decreasing away from ths maximum. π >3 thus 3^(1/3) >π^(1/π)
@misuapolozan3523 3 күн бұрын
This is basic high school math
@bkkboy-cm3eb 3 күн бұрын
3^x > 3x (x>1) ∴3^(π/3) > 3(π/3) = π ∴3^π > π³
@unclesmrgol 3 күн бұрын
It takes seconds to determine that lim(ln(x)/x) as x goes to infinity is zero, and, further, that there's a maximum at e. Since e < 3 < pi, we have everything we need.
@caseywood9781 3 күн бұрын
Pi times three
@kw4093-v3p 3 күн бұрын
I guessed and got the right answer
@juaneliasmillasvera 3 күн бұрын
Uff so tired of this type of videos... a^b > b^a if b>a. END.
@Shyguy5104 2 күн бұрын
@@juaneliasmillasvera not quite right 2^3 < 3^2 and 3>2
@juaneliasmillasvera Күн бұрын
@@Shyguy5104 Mmmm, by bad, so there is not a rule for every case of exponents changing variables? or there is a minimun diference between the two variables since what we can approximate the problem in a generallizate way? =)
@joshualee1595 14 сағат бұрын
@@juaneliasmillasvera well I’m not sure if there is complete generalization for whether a^b or b^a is greater, intuition should tell you that the larger the base number, the more exponent size dictates the overall value of the number. Think 2^3 and 3^2 versus 49^50 and 50^49. Intuition tells me that 49^50 is the larger of the two, but 3^2 is the larger of its pair. The question is where this relationship flips, I’m sure someone has already calculated that limit, and it could be a cool exercise to try out for yourself.
@ChavoMysterio 3 күн бұрын
√[ln(x)]=ln(√x) Domain x>0 ln(x)=[ln(√x)]² Let y=√x, y²=x ln(y²)=[ln(y)]² 2[ln(y)]=[ln(y)]² [ln(y)]²-2[ln(y)]=0 Let z=ln(y) z²-2z=0 z(z-2)=0 z=0 ln(y)=0 y=e⁰ y=1 √x=1 x=1 ❤ z-2=0 z=2 ln(y)=2 y=e² √x=e² x=e⁴ ❤
@HoSza1 4 күн бұрын
Theorem: exp(1) < x < y ⇒x^y > y^x. Corollary: x=3, y=𝜋: 3^𝜋 > 𝜋³. Q.E.D.
@zionfultz8495 4 күн бұрын
Simple to solve really. 3^x = x^3 when x = 3. 3^4 = 81, and 4^3 = 64 therefore 3^4 > 4^3. Therefore 3^pi > pi^3. The only hand wavy thing was 3^x = x^3 when x = 3. You should technically prove this is the only real solution. But since cube root's alternative solutions and logarithms alternative solutions only matter in the complex I do believe it is easy to see there is only one.
@jamesharmon4994 3 күн бұрын
I love your substitution of x for pi. This makes the problem obvious. You can graph 3^x and x^3.
@jamesharmon4994 3 күн бұрын
I love your substitution of x for pi. If you graph 3^x and x^3, it becomes clear that for any value of x greater than 3, 3^x is greater than x^3.
@tamarausher60 2 күн бұрын
This was my solution too❤
@user-cq4st9hh7k 4 күн бұрын
well.. the function f(x) = x^(1/x) is decreasing when x>e. (defferetiate it for proof) hence if e < a < b than a^(1/a) > b^(1/b) hence (a^(1/a))^(ab) > (b^(1/b))^(ab) hence a^b > b^a here 3 > e and pi > e. and 3 < pi. so 3^pi > pi^3 peace to all
@wostin Күн бұрын
Wait, so for every two numbers when one is greater than the other (a<b) then a^b > b^a ??? Or am I just confused?
@tullfan2560 4 күн бұрын
You could use an induction proof. If π was equal to 3, then 3^3 = 3^3. if π was equal to 4, then 3^4 > 4^3 (81>64) if π was equal to 5, then 3^5 > 5^3 (243>125) But π = 3.142... and 3 > π > 4 > 5 Hence 3^π > π^3
@robertveith6383 4 күн бұрын
That is not a correct proof. Pi is not equal to 3.142... That is nor an induction proof.
@robertveith6383 4 күн бұрын
That is not an induction proof.
@tullfan2560 4 күн бұрын
@@robertveith6383 What do you mean pi is not equal to 3.142... That is actually rounded. It should be 3.14159... But, what I'm really saying is that it is between 3 and 4. If it's not induction, what do you call it?
@robertveith6383 4 күн бұрын
@@tullfan2560 Pi, of course, is *not* equal to 3.142... The three dots (an elliipsis), means the digits continue after the 2, which they do not. Pi = 3.141... instead. Induction will involve a base case, and using a variable, for instance.
@tullfan2560 4 күн бұрын
@@robertveith6383 You can see that when the difference between the number and its exponent increases, the value of 3^N - N^3 increases. The difference between the two terms is zero when N=3 and monotonically increases as N gets bigger. As π is greater than 3, 3^π will be greater than π^3.
@AnanthNat 4 күн бұрын
In general, smaller number raised to bigger power is greater than the other way around. 2^8 > 8^2 I hope based on this logic, we can conclude that 3^π > π^3
@zihaoooi787 4 күн бұрын
this is only by heuristic, though. 3^2 > 2^3
@robertveith6383 4 күн бұрын
Your post fails.
@kennethgee2004 4 күн бұрын
yawn 3^pi is large as it is closer to e. this is already forever answered. in any a^b versus b^a if a,b > e and given a<b then a^b is greater. the same is true if a,b <e then if a<b then b^a is greater. if the are on opposite sides of e then it is hard to say and other methods would be required. I conjecture though that if on opposite sides since e is at the critical point whichever the delta is smallest to e for a,b then the one with the smallest delta wins.
@robertveith6383 4 күн бұрын
Write sentences! What you have written are groups of words that do not have capitalized first words or needed punctuation. Do not expect readers to look at your sloppy, lazy, ignorant post.
@hydraim9833 4 күн бұрын
Excellent explanation really. Instantly subscribed, thank you for this video
@sandeepagarwal7387 4 күн бұрын
Excellent. Very well explained... with a simple yet concise flow...
@ikonikgamerz3853 4 күн бұрын
For positive bases & positive exponents, I use the rule of abba or a^b () b^a If a > b then a^b > b^a If a < b then a^b < b^a If a = b then a^b = b^a
@octalbert7280 4 күн бұрын
This rule just doesn't work like if a = 10 and b = 2, a^b < b^a and not the opposite
@SirRebrl 4 күн бұрын
@@octalbert7280 It didn't work for the problem in this very video, either, since 3 < pi and 3^pi > pi^3
@prasoon7916 4 күн бұрын
This is pure trash
@prasoon7916 4 күн бұрын
for positive numbers, to compare a^b and b^a, then if a>b>e, where e is the euler's constant (around 2.71), then a^b < b^a, and if e>b>a, then b^a > a^b, if a and b are on the other sides of e, then it can go both ways and we need to find something else
@robertveith6383 4 күн бұрын
@@prasoon7916-- Write a sentence.
@teodorgalit6878 4 күн бұрын
dude...lol too much. This is good practice for all you know....
@abhirupkundu2778 4 күн бұрын
Bud I solved it orally wdym 95%failed💀💀💀
@jimf2525 4 күн бұрын
Advice: Don’t rattle off the chain rule. Show your work. I gave you a thumbs up expecting you’ll improve.
@allasar 4 күн бұрын
Disagree. I would say most viewers of this channel know how to derive a fraction. Would be tiring to go over each step everytime, already knowing the result of the step. Basically, we skip steps all the time (for instance, we do not need the step by step derivation of lnx to get to 1/x). The question is: which steps do we skip? Personally I would not draw that line at the derivation of a fraction. Most of us have done it so many times, we do not need to see (f'g - fg')/g^2. Nor do we need (1/x × x - lnx × 1)/x^2 as a step, since we already calculated that in our heads before it is written down. There is no clear line, but writing out each step would get old really fast. We are solving an Olympiad question, not filling in a math exam.
@jimf2525 4 күн бұрын
@@allasar We’ll agree to disagree, but not even Michael Penn skips that much math in 1 step. Though I admit he has, but in those cases his videos were quite long. Also, I like that ur willing to fight youtube’s algorithm on length. Content creators are making their videos for too long these days. They constantly stretch one or two minutes of information into 15. So, I probably shouldn’t have complained. 😀
@valeyard00 2 күн бұрын
I'd rather she explained the chain rule than write out ln (a^b) = b ln(a). If you need to spell that out for someone then they aren't gonna understand chain rule off the top of their head either
@betabenja 4 күн бұрын
7:12 sounds like you're threatening me on the street
@fariesz6786 4 күн бұрын
i mixed up NAZ and ZAN again and got confused q3q
@wixom01 4 күн бұрын
I did this in my head.
@joanignasivicente2012 4 күн бұрын
Great!❤
@shehadehaddad8067 4 күн бұрын
But t>0 !!! How you took t=0 as a solution ?!!!
@novidsonmychanneljustcomme5753 4 күн бұрын
In fact it should be t>=0.
@justincase4812 4 күн бұрын
This is not simplification, it's conviluting what was already simplified.
@erlyndavidmarrugoramosestu8855 4 күн бұрын
Yes, in fact, it has two solutions: x=1 and x=e^4.
@swedishpsychopath8795 5 күн бұрын
Or just use induction combined with intuition. Substitute PI with 2. You then have 3 to the power of 2 (which is 9) on the left side. And you have 2 to the power of 3 (which is 8) on the right side. So is 9 bigger than 8? YES! And this is true for all numbers N. If you have PI instead of 2 you know the left hand side will always "win".
@allanflippin2453 4 күн бұрын
I considered this approach, but wasn't able to completely prove it. Suppose we start the sequence with 1^2 and 2^1, the left side is less than. Going to 2^3 and 3^2, the result is greater than as you showed. Working the next few (3^4 4^3. etc) they lead to the left side greater than. Intuitively, if this continues to infinity, the two sides seem to approach equality asymptotically. But how does one prove it? If you're at point N and the left side is greater, how to prove that point N+1 also has left side greater? I couldn't figure a method to prove it that doesn't involve calculus (and likely log as well).
@Patrik6920 4 күн бұрын
@@allanflippin2453 ..did graph this a^x=x^a , at e thers one intersection point, under and over thers two under a<e, a^x is always smaller ofter the secont intersection point the secont intersection point is also always x=a^a after a>e, a^x is always greater after the second intersection point for very lage values of a the first intersection point goes toward 1 and the second is x=a^a tested a between 0 < a < 1 000 000 ..and its always true... edit: actually 0's not tested as 0^0 is undefined, just close to zero values tip: use a log axis scale, makes life easier ;D (a^x becomes a traight line)
@allasar 4 күн бұрын
@@allanflippin2453I think his point is, we inuitively know the answer without proof, so if the question is just "which is larger?" we'd get the points. But I agree, we need proof since intuition can be wrong, and an answer (even a correct one) without proof is dissatisfying.
@allanflippin2453 4 күн бұрын
@@allasar Yup, I am dissatisfied that I can't prove it :D
@samueldeandrade8535 4 күн бұрын
@@allanflippin2453 so, you are trying to prove N^{N+1} > (N+1)^N for N≥3? First, we can prove 2^k < (k+1)! for k≥2. Indeed, for k=2, 2² = 4 < 6 = 3! = (2+1)! ✓ If it is valid for k≥2, then 2<k+2, so we have 2^k < (k+1)! => 2×2^k < (k+2)×(k+1)! => 2^{k+1} < (k+2)! Done. We proved 2^k < (k+1)! for k≥2, which implies 1/(k+1)! < 1/2^k (*1*) for k≥2. Now we prove sum 1/k! < 3 (*2*) for any k. Indeed, 1/0! = 1 1/1! = 1 1/2! = 1/2 1/3! = 1/6 < 1/2², by (*1*) 1/4! = 1/24 < 1/2³, by (*1*) ... Adding everything, sum 1/k! < 1+1+1/2+1/2²+... < 1+2 = 3 So we proved (*2*). Now we prove (1+1/N)^N ≤ sum_k 1/k! (*3*) for any N, with k = 0,...,N. Indeed, expanding (1+1/N)^N using the binomial formula (1+1/N)^N = sum_k C(N,k)(1/N)^k with k=0,...,N and C(N,k) = N!/(k!(N-k)!) So, fixing k, 1≤k≤N, we have the term C(N,k)(1/N)^k = (N!/(k!(N-k)!))(1/N)^k = (N(N-1)...(N-k+1)/k!)(1/N)^k = (N/N)((N-1)/N)...((N-k+1)/N)(1/k!) = 1(1-1/N)...(1-(k-1)/N)(1/k!) ≤ 1/k! because for each factor 1-j/N, j=0,...,k-1, 1-j/N ≤ 1 So we proved (*3*). This means we have, for any N, (1+1/N)^N ≤ sum_k 1/k! , by (*3*) < 3 , by (*2*) proving (1+1/N)^N < 3 for any N, which implies (N+1)^N < 3N^N (!!) Finally ... for N≥3, we obtain the inequality, (N+1)^N < 3N^N , by (!!) ≤ N×N^N , by 3≤N = N^{N+1} Done. That's a way.
@AzttcAzttc 5 күн бұрын
Here 95 % refers to kindergarten students 😂
@romank.6813 5 күн бұрын
Compare these two: sqrt(6)^pi and pi^(sqrt(6)).
@-wx-78- 5 күн бұрын
Коварство запредельное: √6 и π лежат по разные стороны от e. Хотя можно и усугубить: √7<e. 😉
@romank.6813 5 күн бұрын
@@-wx-78- Неа, лучше (е-1/е)^π и π^(е-1/е)
@-wx-78- 5 күн бұрын
@@romank.6813 Месье знает толк в извращениях. 😉 P.S. В который раз вижу под русским текстом “Translate to Russian”, сподобился нажать - и вместо объяснения про константу Эйлера получил замену «Неа» на какой-то “her”.
@lawrencejelsma8118 5 күн бұрын
By squaring and exponentiating results you are expanding your domain to solve a domain questionable change during the solution of the problem. In square root exoressions considered a positive solution by squaring a result introduces the negative root as an extraneous solution "liberal mathematicians" say is invalid. In Physics and Electrical Engineering mathematics "extraneous solutions" mathematics is an applied concept because of direction of vectors notations for vector math. Because a "-" result can happen with each "+" result in vector math where a square root was initially considered a positive, Physics of the sciences in mathematics define rules such as "vector right hand rule" to define the vector's direction even though the sauare root took a square root of a positive magnitude direction. Further academics of teachers explaining vector "right hand rule" for the vector "cross product" further answers this thought process more in the "mathenatical sciences!" ✓(lnx) = (1/2)(lnx) is (lnx) = 1/4((lnx)^2) by squaring the left hand side and the right hand side of the equation "mathematical sciences" routinely perform to contain an extraneous solution or root for further analysis. If lnx doesnt equal 0 such as in the ln(1) case then 1/4ln(x) = 1 or ln(x) = 4 or x = e^4 other solution. x = 1 or x = e^4 both solve the initial given equation. In sauaring and exponentiating we know x = 1 or x = e^4 and since 1 is not equal to e^4 we know x is one of those solutions or the other. Because both solutions silve the given equarion there were no extraneous solutions that didn't solve the given equation. The graph of the two curves would have intersected twice in all domain of x (the domain expanded by the exponentiated process).
@lawrencejelsma8118 5 күн бұрын
Note! All spelling errors were created by my app that doesn't grammar spell error correct like "Google" and "Microsoft Office!". KZfaq is not a U.S. company so spell checking is not in its helping programming. I can easily write a spell check program, personally, but KZfaq has its flaws to allow poor grammar into sentences.
@honestadministrator 5 күн бұрын
i √8 * i √9 = -6 √2
@hangthuy458 5 күн бұрын
(-8×-9)^1/2=72^1/2=(36×2)^1/2=6×2^1/2
@user-yc3sh1ij8b 5 күн бұрын
RD sharma problem😅
@novidsonmychanneljustcomme5753 5 күн бұрын
Correct solution (aside from the wrong domain as already pointed out in other comments), though sorry, but I'm sick of these clickbait video titles. 🤦♂️ Why not just name the mathematical problem itself in a creative way (e.g. "logs & roots in one equation")? And do you have any statistical sources for your "95%" claim? It might make some people feel better thinking they are part of the 5% after they got the correct solution, but not me. I solved it by myself as well before watching your solution, but that doesn't make me proud in this way, but I rather think this problem definitely ain't that hard as you claim. So I don't only think that these "95%" statistics don't exist, but even if they did, they would be wrong. Not meant to offend you, but I hope you get my point. Have a nice day. ✌️
@honestadministrator 5 күн бұрын
√( ln (.x)) = (1/2) ln ( x) ln ( x). = 4 x = e ^4