A Nice Olympiad Exponential Problem.
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- Опубліковано 25 сер 2024
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How to solve this Exponential math Olympiad problem
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The logarithm of the number a on the base of b is by definition the number in which you need to raise b to get a. Therefore, when you saw that 2^x=5, you could immediately write an answer with a logarithm. No extra frills.
Shoot.. the problem attracted me to guess. If x was 2; then 68 was smaller than 130. If x was 3 then 520 was greater. So I guess around 2.5. 😂
@Repent and believe in Jesus Christ I don't think Jesus would've solved the equation for me
@repentandbelieveinJesusChrist2 i believe John was Asian dude
Yeah. Log a b is like a mark for “how many time of value a would get me the value b”.
the extra steps are needed for a rigorous proof, if u need just the answer you would be correct
The trick to solve this one is to get to (2^x)^3+2^x=130 and notice that 130 is a decently small number for a polynomial of form y^3+y=130. Testing the first few integers will get you to 5^3+5 = 125+5 = 130. Which implies that 2^x = 5. Solve for x by taking log_2 of both sides and you get x=log_2(5).
Note that you need to prove that 2^x=5 is indeed the only solution. Best way to do that is to prove that f(x)=x³+x-130, x>0 is strictly increasing, which is obvious, but necessary. Saying this for the younger audience.
My 15yr old daughter worked it out faster. Using less stages.
@@lesleyrobertson5465not a competition but my 2yr old son did it in 5s using only 1 step
@@lesleyrobertson5465 Nice one, but we didn't ask :)
@@lesleyrobertson5465 Then show us
8^x+2^x=130
(2^x)^3+2^x-130=0
y^3+y-130=0
y^3-125+y-5=0
(y-5)(y^2+5y+25)+1(y-5)=0
(y-5)(y^2+5y+26)=0
y^2+5y+26=0
a=1, b=5, c=26
☆=5^2-(4×1×26)
☆=25-104
☆=-79 (solutions not possible)
y-5=0
y=5
2^x=5
x=log_2(5) ❤🧡💛💚💙💜
I thought the very same
Basically how I solved it as well.
@@jcb3393 using the difference of cubes theorem and factoring by grouping.
if we Substitute the xvalue in equation . The ans is 130.018 so wrong question. Sure
Atleast you can solve it
This is a horrible method, no one is going to guess to break up y into 25y and -26y. You could have just used rational root theorem or even vietas formulas to find the roots for y. And even when your checking the solution there is no need to approximate x for 2^x=5, because when you plug it back into the original equation you obtain 2^3logbase2(5) + 2^logbase2(5), which gives 130 exactly. Since the 3 gets moved back to the power to obtain 125 and then using definition of the inverse function. Since logbase2(x) is the inverse of 2^x.
5% of 18-year-olds in Turkey do this
To make it worse, Math Olympiad doesn't allow calculators, so there's absolutely no point to approximate the value of log_2 5
Vietas!
I thought im the only one feel this method was absurd 😅
I don't dislike that method , but I didn't understand how he got it.
when checking, no need to get value of log2,5. since 8^(log2,5) = (2^3)&(log2,5)= 2^log2,(5^3) = 5^3 = 125
I have just realized that my life without this kind of math is so nice.
U r funny
That's what I am literally studying
Pain and suffering for a simple x²+x+1
Only just ?This leaves my head in such humility
I can't believe I watched this whole thing and nodded along as I told myself "I know exactly what this guy is talking about"
Guys, this is olympiad. It is not just using a calculator. Please, do consider the effort he has made so that we can learn more.
Well if it is more complex it could be 8^-2 + 2^7=130 or 8^2.25 + 2^1=130 if there are two different x values instead of the normal 1. It does not say solve for x. It asks how to solve.
Just to summarize and clarify two comments already made,
note that: log_b(b^a) = a because log and exponentiation are inverse functions when using the same base b,
and this one fact can be used to both help solve and verify the answer as follows:
1) 2^x = 5 --> log_2(2^x) = log_2(5) --> x = log_2(5) which of course is the answer.
2) 8^x + 2^x = 2^3^x + 2^x = (2^x)^3 + 2^x = (5)^3 + 5 = 125 + 5 = 130 which of course verifies the answer.
Alles sehr willkürlich ...
@Repent and believe in Jesus Christ fok off with your god. Only trouble religion has given us so far.
great verification, much better than one in the original clip (with approximate numbers)
I did maths at university for 3 years but have not used it for many years
I enjoy your videos, takes me back
Maths is fun , you have a passion so thanks and ignore the negative comments
I can see now why the engineering of buildings, bridges and rollercoasters can go so terribly wrong.
In the last step showing the alternate base, why not use log base 2 instead of log base 5. That would immediately give you x(1)=log base 2 of 5. It shortens the calculation.
这疑似在骗流量，把一个简单的题用二次替代法绕了个大圈子，130分拆也无必要，可以取个简单数，最后答案也不是个整数，无人能人工计算非特殊值的对数。
Not good because that would immediately give you an answer of 2.32. 😅
@Repent and believe in Jesus Christ All praises are to God for what we have.
Wasnt good at maths when i was young, but i appreciate the different methods used to come to the final answer.
The equation suggests the value of X is not an integer and trying few numbers we realized the value of X should be a real number between 2 and 3. As the other side of the equation is an integer, the left side must be integer too. So, by evualuating the smallest term 2^x by x = 2 yields 4, the next integer is 5. So, the other term should be 125. If we equal each power to these integers accordingly: 8^x = 125 and 2^x = 5, applying logarithm rules and solving for x. The value is x = log(5)/log(2) or 2.3219... which turns to be equal to log(125)/log(8). I solved this by assuming 8^x and 2^x were both integers which in other cases might not be true.
2 ^3X and 2^X is equal to 2^7 and 2^1
Not exact, but the right hand side and left hand side are all at base 2
A quick glance means X must be between 1 and 3, and (not precise) but 7 ÷ 3 is 2.33333
solve t³ + t = 130 = 5*26; you see that t = 5 ( function is always accending), so x = log_2(5)
👏👏
Solved it numerically in my head.
1) No integer x satisfies equation.
2) The value of x must be between 2 and 3.
3) It’s likely closer to 2 than 3.
4) Evaluate a few cases (or perhaps just one) and find x is appropriately equal to 2.3.
Not exact like log2(5) but certainly expedient.
In multiple choice question:
A. 2.30
B. 2.31
C. 2.32
D. 2.33
Which one will you choose😁🤣
@@agusludin7947 Four possibilities in the second decimal means at most two evaluations to compute the correct multiple choice answer.
Really good videos,Thanks.Good luck
Thanks, you too!
Solve 5+5+5=550 with only one dash (stroke) there are 2 solutions
Watching this video reinforces my belief that I’m just not into math…..this kind of question never fails to give me headaches….
130 is 1000010 in binary. Done.
Not a great solution in general; for this specific problem, magnificent.
令A=2*x,則A的3次方=8*x
原式：A*3+A*1=130,則A（A*2+1）=130
A=5，則2*x=5,x=log5/log2
或x=log以2為底的5
Good method when you have to solve only one question in 3 hours...
Avec : X=2^x, (i.e: logX=xlog2),l'équation donnée peut s'écrire X³+X =5³+5,d'où X=5, log5=xlog2, x=log5/log2.
(les amis, je vous laisse le soin du détail).Ceci pour dire qu'il faut en faire UN JEU plutôt qu'une laborieuse démarche scolaire .
▪︎observation ▪︎recherche d'astuces.
wow you are so smart
Math is sexy when you know what you are doing, and terrifying when you don't know where to start
I agree sir
Amazing method being a math student these are very simple to understand
Glad it was helpful! Thanks ❤️🙏❤️
With y=2^x, we have the equation y³+y = 130, or y(y²+1) = 130. If this have an integer solution for y, then y | 130, y=5 seem appropriate, so let's check: 5²+1 = 26, 5×26 = 130, and this is an answer. x = log₂ 5.
y³+y = 130 should have other two answers: y³+y-130 = (y-5)(y²+5y+26) = 0. The other answer: y = ½( -5 ±√{25-104} ) = ½( -5 ±i√79 ) = √26 e^{i something} there would be numerably infinite complex solutions for x, and x = log₂ 5 as the only real solution
I get that he’s showing a process (though it’s seems needlessly complicated) when (for this example at least) it’s so simple - the exponent for the 8 can only be 1 or 2. So take a guess and start with 2. 8x2=64 130-64=64 2x2x2x2x2x2=64
It’s like 7 seconds in your head - no paper required. Maybe an extra 5 seconds if you guess 8 exponent 1 and bring the 2 exponent up until it adds up to 130, but as soon as you see that 64x2=128 you realize that it must be 8 exponent 2 (cause it can’t be 1 or 3) plus 64 (which you already got on your way to 128). So you take one exponent off your 2 and bring your 8 exponent up to 2 to match your other 64 and you’ve got it. Why make something so simple so complicated?
Большое спасибо. Очень интересно и познавательно.
Thanks mate for interesting explaination.
8^(log 2 of 5) = (2^3)^(log 2 of 5) = (2^log 2 of 5)^ 3 = 5^ 3 = 125
No need to use approximate values.
Thanks for that!
Yeaaah. I have must check that solution is here.
At Y^3 + Y = 130 I would take a few guesses at y. Try 4, gives 68 too low. Try 6 gives 222 too high 5 gives 130 so 2^x is 5
↓ *Japanese high school student's way*
8ᵡ+2ᵡ=130
y³+y-130=0 (y=2ᵡとおいた)
y³-125+y-5=0
(y-5)(y²-5y+25)+(y-5)=0
(y-5)(y²-5y+26)=0
y²-5y+26=0の判別式をDとおくと
D=(-5)²-4•1•26
Так можно до ,,черной дыры" дорешаться...
I had X at 13 . (8x13)+(2x13) = 130. So much for remedial math skills
Yes
Very good
Thanks
Let y = 2^x, then y^3 + y = 130. y(y^2+1) = 130 has only one real root. By the rational root theorem, the possible rational root is an integer,
and it divides 130. 130 = 2*5*13. Trying these can easily determine that y = 5 is the solution. x = ln(5)/ln(2).
While watching this video i had an urge to go back and continue college after 25 years dropping out 😥
That's sad man
Oh I still have no interest in going back to college. I also dropped out, and this video reminds me of why.
@@yummygoy5138 The saddest part my mom lived the dream of me going back to finish college until she passed away 7 years ago.
And that was the last reason that could make me actually consider going back.
This looks extremely complicated! But I sucked at college algebra and logarithmic formulas drove me crazy. I’m not one of those people who love math and find it fun. I look at all those log formulas and it seems like a foreign language. I just couldn’t get it.
Да, представить 130 в виде 26*5 с ходу не получится. Клёвая задачка.
It’s nice how any computer programmer / engineer / scientist can say : it’s just 130 in binary and get an answer literally in 2 or 3 seconds. Now the fun part is searching for non integer solutions
Of course, there are an infinite number of complex solutions to this problem when non-principle branches are admitted.
That is why I say modern mathematics are bullshit!
Attack intuitively. Let 2^x equal a, then a^3 + a = 130. By inspection, a = 5. Therefore x = log 5 / log 2 follows immediately. The rest is trivial calculation.
This is how I was taught to solve cubic equations. Start off with some guesses. Didn't take long and you've explained it in 1/(26y-25y)🤷♂️ the time he tried to explain it
Correct
Это решается с помощью теоремы Безу. Сначала замена переменной, а потом сама теорема.
It is very good it is excellent
There is a easy method for factorisation, which is called zero method. You go on putting y equal to 0,1,2,....andso on until the lhs vanishes. In this case y=5 makes lhs zero. Hence y-5 is a factor. Rest is easy.
Хорошо.Для тех кто не знает формулу" разница кубов."
Thank you for the video.
You're welcome
The answer to y^3+y=130 is obviously y=5 by inspection. If you want to make these a challenge, you have to pick problems that don't involve small, whole numbers...
You can also solve the problem by using remainder theorem.
True, I don't know why he did so much hassle
Better method is after coming down to the equation y^3+y= 130, try to substitute nos. Y cannot be even, so put y= 1,3,5. y= 5 is a solution. Now, divide y^3+y-130 with y-5, and you get a quadratic equation which can easily be solved further. The put y= 2^x and here is the solution.
I am useless at maths but even I spotted that he had a quadratic that even a cheap Casio will solve
you could even stop when u had 5 as an obvious solution. you need to make sure that's the only solution tho. Which it is, because we have an increasing function
For me the easiest method is adding together the 8 and the 2 then you get 10 . 13 divided by 10 is 13 and you get the answer
I like your solution
Ничего не понятно, но очень интересно !
А ещё говорят - математика - точная наука! Почему тогда ставится значок "поимерно"? Я решил этот пример без всяких log и х 8 в 10 степени + 2 в 20 стерени = 130. Это же надо так мозг людям запудрить!? 5 лет в школе изучали алгебру, в жизни так нигде и не пригодилассь в отличии от геометрии!
130=13.10=26.5=(25+1).5=5^3+5; T^3+T_130=0, (T=2^x >0); T^3-5^3+T-5=0; (T-5)(T^2+5T+5 +1)=0; vì (T^2+5T+6>0); T-5=0; T=5 =2^x;
I did as follows:
8x + 2x =130
10x (add 8 and 2 together) =130
130/10 =13
X =13
So...
(8x13) + (2x13) =130
104 + 26 =130
In the last part simply use logarithmic formula
8^(log5 base 2)+2^(log5 base 2) = 2^(log(5^3) base 2)+2^(log5 base 2)=5^3+5=130 .
If I may use my cell calculator, I find easy, X = 2,322
Prove: 125,018 + 5 = 130,018
Как я любила решать подобные примеры в школе!
💯
Кошматерный сон для меня))), зато зиготы, гаметы и прочие апендициты👍, одуванчики с пестиками и амёбы🥰!
O desenvolvimento da equação foi brilhante , porém no final ele complicou o resultado com tantas propriedades de log. Bastava calcular direto 2^x =5 ; x=ln5/ln2 ; x=2,3219281 .
Very detailed explanation ! Thanks for sharing ……..
You are welcome!
Каждый ученик в моей школе мог решить эту задачу! Элементарно.
Кавказская хвастливость?
130 = 128 + 2 = 2^7 + 2^1 might be a good way to start.
I’m not sure I understand why that would help
All I did was add the eight and the two together and I got 10 and divide 130 by 10 so x=13
The best way to make kids hate calculus! Bravo 👏
Where TF ... Did he use calculus ?
It's simple algebraic factorisation and simplifications including logarithms.
This is why I majored in history 😂
I really believe in the trial and error method for this one. It literally took me 90 seconds. X=2 was to low. X=3 was too high. A few shots in the middle and after 5 tries X= 2.321. Sometimes it's more efficient to just use logic.
Which would give you zero points on any math exam I ever took.
@@StCreed You never took a multiple choice test then huh.
That is not a proof !!!!
@@ChasOnErieI didn't see that it was a proof. I seen an equation, I solved it.
This is for an Olympiad, who’s to say you didn’t look over at someone’s work to get the answer…. that said, your approach would work just nicely in a game show
Just apply log both side logx= log130/4, x=log-2(5)
Very interesting method
I would have tried it differently but thata the beauty of math. There's always another way
Glad you liked it
A Nice Olympiad Exponential Problem: 8^x + 2^x = 130
8^x +2^x = 2^3x + 2^x = 130 = (5)(26) = (5)(25 + 1) = (5)(5^2 + 1) = 5^3 + 5
Convert the exponential base number 5 into 2 using logarithmic math:
Let 2^n = 5, n = log5/log2 = 2.322; 5 = 2^2.322
8^x +2^x = 5^3 + 5 = (2^2.322)^3 + 2^2.322 = (2^3)^2.322 + 2^2.322
= 8^2.322 + 2^2.322; x = 2.322
Thanks 🙏
A very elegant solution. 👏
Почему после введения замены не применить деление на многочлен
(у - 5)?? Очень долго шли к квадратному уравнению. И потом, что оно не имеет корней, видно уже из скобки - сумма всех положительных слагаемых не может быть = 0)) И очень, ну очень долго решаете показательное уравнение. Там 2 строчки по определению логарифма.
За рубежом простые методы решения уравнений попросту не изучают. Нет необходимости. Именно поэтому типичная задача подготовленного выпускника школы отнесена к олимпиадным.
Сложно о простом.
Great sir
y^3 + y is a strictly increasing function for positive y and 5 is a solution to y^3 + y = 130, therefore it is the only valid one. No need for most of the algebra.
How can I apply this method to a similar equation 8^x + 2^x = 131? Or 8^x + 2^x = 132?
For the general case you would reduce it to the cubic in y, as per the video. You could then try Cardano's method to solve the cubic.
@@ashleyhubbard8854 Exatly. The above shown way only works, if you have known/guessed a solution beforehand. What I wanted to say is, that the method works only for RHS 130 and not any others.. It is not generally appliable...
You can solve the problem:
1) to guess a solution and reduce to the quadratic equation.
2) to apply rational root theorem, and try all divisors of 130, if they are a solution. I.e. y = +-1,+-2,+-5,,+-10,+-13,+-26,+-65,+-130. And again reduce.
3) to use a variant of cardano formula
4) to use a numerical software
...
Or to ask WolframAlpha "reduce y^3+y-130".
@@lubosdostal8523 You are right. The 130 is obviously a contrived number.
ОХРЕНЕТЬ !!!
Its very beautiful math problem
Thanks and welcome!
Nice solution,
I'll just tell my approach to this problem.
Here' my 2 cents:
8^x+2^x=130
2^3x+2^x=130
Take 2^x=t for convenience
t^3+t=130
t(t^2+1)=130
Factorizing 130 you will get,
130=2×5×13
Now just do hit and trial and u will get t=5
2^x=5
X=log5base2
you are very good at explaining things
Thanks and Welcome!
I solved this in 20 seconds by just trying a few numbers. 8x13 = 104 2x13 = 26 26+104 =130 So "X" is 13. Pretty simple
This is reason why indians always in top of CEOs at big companies
Excellent
Thank you! Cheers!
a³+a=a(a²+1)=130
a²+1 might have 26 because of 130. a²=25 a=5 then we control the equation, it's true( what a surprise ). a is equal to 2*x. So x is equal to log2*5. That's simple
Помните анекдот:
- ........
- Сколько будет дважды два?
- 4
- Неправильно!
- 5
- Неправильно!
- 6
- Неправильно, но уже лучше.
- А как правильно?
- Дважды два будет приблизительно 7 или 8.
Good one
Very interesting ❤
Thanks and Welcome 🙏❤️🙏
So I got eight boxes with 432 million pencils in each box plus two boxes with 632 billion and each box. Which pencils were the first, middle and last, handled during the packaging.
Totaling 3 pencils only.
Here's another one I have a chicken that's 18 years old when did it lay its 300th egg.
Valu
I RESOLVED SIMILARE IN MY IDIOM SPANISH. THANKS
hello
when you have y^3+y=130 the problem is finished since 130 =125+5=5^3+5 and y=5
2^x(2^2x+1)=130
5(5^2+1)=130
2^x=5
×=log_2 (5)
substitution : y= 2^x; then: y^3+y = 130 = 125 +5 = 5^3 + 5 => y = 5; back to substitution: x = (ln5/ln2)
На банальный подбор 2 минуты ушло, нахкя так засорачиваться🤣
I would just enter X on my calculator and iterate different values til finding an answer close enough
Good, i like this.
Ich komme mir nun freilich blöd vor, zu bemerken, dass ich mir blöd vorgekommen war, als ich sah, dass es ja nun doch etwas gedauert hatte, ehe ich gesehen hatte, dass das ja ganz einfach war und nun so schön, wenngleich nicht rational, aufgeht. 11 Minuten hat das aber nicht gedauert. Ein recht hübsches Molekül. Ich danke.
Thanks you teacher
Bravo ✨✨✨✨✨✨✨✨👍🏼
Thanks! Your welcome 👍👍👍
so easy
8^x + 2^x = 2^(3x) + 2^x = 2^x (2^(2x) +1) = 130.
So we have product of two numbers that gives us 130, and one of them is the square of the other + 1.
Candidates:
2, 65 -> 65 is not 2^2 +1
5, 26 -> 26 is 5^2 +1
10, 13 -> 13 is not 10^2 +1
Thus, 2^x = 5, thus x = log_2 (5).
Of course this eliminates other answers in theory, because 130 could be a product of not only integers, but it's very quick.
Funny