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Mexico - A Nice Math Olympiad Exponential Problem 

LKLogic
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Maths Olympiads are held all around the world to recognise students who excel in maths. The test is offered at many grade levels and provides them with numerous possibilities to win certifications, awards, and even scholarships for higher studies.

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15 июн 2023

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Комментарии : 537   
@honeytgb
@honeytgb 8 месяцев назад
This is why I'm going back to watching cooking videos.
@nitdiver5
@nitdiver5 4 месяца назад
Or funny cats
@Aman-nk5uq
@Aman-nk5uq 4 месяца назад
Come on. Everyone should know how to solve this
@xtranub8792
@xtranub8792 3 месяца назад
​@@Aman-nk5uq he means it's too easy
@Fingolfin3423
@Fingolfin3423 Месяц назад
Most mathematicians probably can't cook worth a damn. Most people can't.
@dr.komiemmanuel9748
@dr.komiemmanuel9748 10 дней назад
😂😂 I don't blame you
@andreadanieli6192
@andreadanieli6192 10 месяцев назад
At the end you can simply apply the definition of logarithm... 2^x = 5 ---> x is the power to wich the base 2 must be raised in order to obtain 5 so you can write: x = log2(5)
@tintiniitk
@tintiniitk 8 месяцев назад
No, she has to do a 5-minute thesis on it (to go from 2^x=5 to x=log5 to base 2, otherwise her students might not understand it.
@andre.moonlight
@andre.moonlight 8 месяцев назад
@@tintiniitkwe learned it as logba=e where e is the exponent b is the base and a the “answer”. its very helpful
@logminusone1272
@logminusone1272 10 месяцев назад
I feel the quick way to approach this problem is to recognise that we have a polynomial where both y^3 and y exist. This should ring a bell that we should find some cubed number somewhere. We can write 130 = 125 + 5 = 5^3 + 5. That's it. Immediately we get y^3 - 5^3 + y - 5 = 0, or (y -5)(y^2 + 5y + 26) = 0, where y = 2^x. Here onwards, it is simple.
@josephazar3516
@josephazar3516 8 месяцев назад
(8×15)+(2×5)=130
@noobLOL77
@noobLOL77 8 месяцев назад
thats what i did, i got hung up because i did log(5)-log(2) instead of log2(5)
@IvyANguyen
@IvyANguyen 8 месяцев назад
Wow. This was way easier! I should've just thought through some numbers and would've hit it just by thinking of those 1st few perfect cubes.
@NikhilSingh-ge8yu
@NikhilSingh-ge8yu 8 месяцев назад
y=2x , find the value idiot
@DantesInferno61
@DantesInferno61 8 месяцев назад
She made it way harder to solve than she had to.
@georiashang1120
@georiashang1120 9 месяцев назад
130=26×5=(5^2+1)×5=5^3+5; 8^x+2^x=(2^x)^3+(2^x); 2^x=5 My high school math teacher used to tell me,to understand an equation better you need to dis-simplify the brief side other than to simplify the complex side.
@NITian.Navneet
@NITian.Navneet 8 месяцев назад
Wowwww😮😮😮
@July-gj1st
@July-gj1st 8 месяцев назад
Nice! I did it like this as well or, similar.
@StarLord1994
@StarLord1994 7 месяцев назад
Yeah, it just depends. A lot of times that advice will work, but not always.
@zhenyuzhai4098
@zhenyuzhai4098 3 месяца назад
Yours can't guarantee there is unique root.
@georiashang1120
@georiashang1120 3 месяца назад
@@zhenyuzhai4098 I know it should be unique in this case
@shannoo7
@shannoo7 8 месяцев назад
Me given IIT 20 yrs back and working in IT from 15+ yrs still willing to learn this so that I can teach my kid. Feels like back to square one. 😂 happy learning guys.
@daakudaddy5453
@daakudaddy5453 8 месяцев назад
Looks like you're stuck in a loop in life.
@krishmav
@krishmav 8 месяцев назад
Don't teach to kids whatever that's not useful. Times are changing rapidly. Pretty sure you never used this in your working life.
@Aman-nk5uq
@Aman-nk5uq 8 месяцев назад
*for 15 years
@ankurmondal3220
@ankurmondal3220 7 месяцев назад
​@@krishmavthese type of outside the box question are what asked in sof olympiads
@Smart_Soham
@Smart_Soham 7 месяцев назад
​@@ankurmondal3220so you are studying fkr sof olympiads, go and give the real IMO Make sure you don't piss in your pants
@mangofan01
@mangofan01 11 месяцев назад
Instead of using decimals you could have used more logarithmic identities when you were checking. I think that would have been cleaner.
@RedPardo
@RedPardo 9 месяцев назад
Came here to note that.
@manojpadmanabhan2615
@manojpadmanabhan2615 8 месяцев назад
Exactly.. 2^(log 125/log 2) = 125 and so on.
@antoniou.1158
@antoniou.1158 8 месяцев назад
​@@manojpadmanabhan2615 let's be nice with our math mates
@PlasmaFuzer
@PlasmaFuzer 8 месяцев назад
Change of bases can be used to simplify the final expression 8^log_2(5) = (2^log_2(5))^3 by associativity and commutativity =5^3 since x^log_x(a) = a =125 And the same with the other term, so 125+5 = 130
@danieldudley5852
@danieldudley5852 7 месяцев назад
I agree, the finish could have been done a bit more elegantly with out finding the decimal approx.
@wiseview1444
@wiseview1444 9 месяцев назад
All these problems seem a little too easy for olympiads
@estefanocrespo7930
@estefanocrespo7930 8 месяцев назад
Are you sure?
@human18711
@human18711 8 месяцев назад
​​@@estefanocrespo7930yes these are like middle school problems for us
@ritsuboii2130
@ritsuboii2130 8 месяцев назад
​@@estefanocrespo7930im not him but yeah this is pretty easy for olympiads because i can even answer it
@alexcwagner
@alexcwagner 8 месяцев назад
Maybe it's the special olypiad
@gauravbasu98
@gauravbasu98 8 месяцев назад
😂😂😂😂
@IoT_
@IoT_ 8 месяцев назад
FASTER APPROACH : 8^x+2^x is always increasing function, hence one root only. Setting x equal to some numbers, we realize that x is between 2 and 3. Consequently,y is between 4 and 8, but y is obviously less than 6 by analyzing cubic equation. So y is between 4 and 6, try 5 and we get that y=5. Everything can be solved in mind.
@akshatgour8981
@akshatgour8981 8 месяцев назад
I did it like this... 2^(3x)+2^x=130 Let 2^x=y Then y³+y=130 By observation, y=5 Hence 2^x=5 X=log(2)5, ie log 5 base 2 Overall, this problem was on the easier side if it's from olympiad, as I and my friends (Indian high-school students) were able to solve it pretty easily 😅
@digitalmarketing8230
@digitalmarketing8230 8 месяцев назад
Hello. I am a mathematics enthusiast from Nigeria and I really want to increase my knowledge in mathematics. I would like to connect with you. Do you mind?
@anthonyl3440
@anthonyl3440 8 месяцев назад
Same thoughts here. I literally solved it in mind using the similar way as you in 1min
@m3zuss
@m3zuss 8 месяцев назад
Yeah but try to solve it when u were 13 or 15.
@Quiablo
@Quiablo 8 месяцев назад
Damn we really dont need a 8 minute to solve this one. Nice observation.
@shashankkhatri5523
@shashankkhatri5523 8 месяцев назад
people @13-15 in india can solve this way faster compared to the older folks in india btw@@m3zuss
@kathrynstemler6331
@kathrynstemler6331 8 месяцев назад
I remember fondly a time where maybe, just maybe, I might have had some idea, but I haven’t used anything beyond grade 9 algebra in 20 years.
@schlingel0017
@schlingel0017 8 месяцев назад
Incredible how much effort you put in this. 130=5³+5¹ and this is identical to your y³+y. Therefore y = 2^x = 5. => x = log2 (5). Easy.
@alexeivalyaev5416
@alexeivalyaev5416 8 месяцев назад
It’s just to prove that there is no other solutions with real values
@daakudaddy5453
@daakudaddy5453 8 месяцев назад
You also have to check for other possible solutions. Duh!
@adriendebosse6941
@adriendebosse6941 8 месяцев назад
y3+y is stricly increasing, thus has only 1 real root, no need to complexify all what he did, like he did with all his base 2 and 5 logarithms. Just write 2^x=5 equals to x = ln(5)/ln(2), it's the basics of logs and exponentials
@josephmiller38
@josephmiller38 9 месяцев назад
Just subtract 130 from both sides and solve for the x-intercepts of the equation 8^x + 2^x - 130 = y. When y= 0 the graph intercepts the x axis.
@one2too402
@one2too402 4 месяца назад
I loved my Math's Teacher Sir A Hameed Wayn, in Metric Class ... And after him, you are the 2nd one, whom I would like to praise .. Since my First Teacher changed the School , and just for Mathematics I followed him to that school. From there, you can see my love for Mathematics . Liked your V-Log ... Though i dont know your name .
@EZ4U2SA.007
@EZ4U2SA.007 8 месяцев назад
This is why I know anatomy so well!
@arcticwolf49
@arcticwolf49 11 месяцев назад
We will also have complex values of x... Since we have a cubic polynomial y^3+y-130=0, it will therefore have 3 roots out of which 1 is real (log5/log2) and other two are complex Other two values of x hence will be: x=log(5+- isqrt71)/2/log2
@NicholasOfAutrecourt
@NicholasOfAutrecourt 9 месяцев назад
The imaginary solutions were extraneous. That's why they were explicitly discarded.
@rickymort135
@rickymort135 8 месяцев назад
​​@@NicholasOfAutrecourtextraneous or extra anus?
@lucianopena3806
@lucianopena3806 8 месяцев назад
So hard. Congrats. Several math concepts 👏👏👏👏👏
@dimchodimov2424
@dimchodimov2424 6 месяцев назад
For all functions of type f(x)=ax^3+bx+c the equation f(x)=0 can have only one real root in case of a>0, b>0, because f'(x)=3ax^2+b>0 and therefore f(x) is monotonically increasing. Also, if the equation f(x)=0 has rational roots in the form p/q, p is a divisor of c and q is a divisor of a. If a=1, the rational roots are always whole numbers. One can immediately see that in the specific example f(y)=y^3+y-130=0 , y=5 is a solution, because we first check the whole divisors of 130 (+-1, +-2, +-5, +-10, +-13). Try to solve x^7+x^5+x^3+x=170 with your method....No chance. With above, you show that x=2 is the only root in just 2 rows.
@IvyANguyen
@IvyANguyen 8 месяцев назад
I do not yet have the mathematical experience to have come up with that line of thinking to jump to thinking of the factorisation of 130 then rewriting as factor by grouping. I got to the step where I substituted u = 2^x but had no idea that was actually the way to proceed. I was stuck at what to do now with the 130 as I had no obvious way to factorise u^3 + u - 130 = 0.
@kaleecharan495
@kaleecharan495 7 месяцев назад
In solution checking a^(log c to base a) can be written as c^(log a to base a) . So 8^(log 5 to base 2) can be written as 5^(log 8 to bas 2), which is 5^3 and 2^(log 5 to base 2), is 5^(log 2 to base 2) which is 5 5^3+5=130
@pintub37
@pintub37 11 месяцев назад
It's too complicated, I mean after getting 2^x = 5 we get in accordance with the definition of a logarithm x = log2(5) where log2 is a logrithm with base 2. All following calculations in the video are unnecessary.
@ildar.ishalin.chelovek
@ildar.ishalin.chelovek 9 месяцев назад
Yeah, exactly my thoughts: just use the definition!
@intigamaghamuradov4727
@intigamaghamuradov4727 8 месяцев назад
It was painful to see how she derives x after getting 2^x = 5 (((
@random_youtube_user
@random_youtube_user 8 месяцев назад
Thanks you so much for refreshing my memory from 10 years ago, Now I wish I have continued on the math field instead
@ariniulfah4885
@ariniulfah4885 6 месяцев назад
Confuse the number 26y - 25y, can you explaine it!
@nielsstobbe2646
@nielsstobbe2646 5 месяцев назад
First observe that 8^x=(2^3)^x=(2^x)^3. Then set 2^x=y and you get the equation: y^3+y=y*(y^2+1)=130. A few trials gives y=5 and thus x=ln(5)/ln(2)=log(5)/log(2).
@kanguru_
@kanguru_ 8 месяцев назад
Let a=2^x, then a^3+a=130, so a=5 is a solution, then x= ln5/ln2. The other solutions can be found by factoring out (a-5) and solving the quadratic.
@CiarcellutiAcademy
@CiarcellutiAcademy 9 месяцев назад
The longest way to do that equation.
@hardtimes2597
@hardtimes2597 8 месяцев назад
4:50 Redundant. By definition of logarithm, it is the value of the exponent to put on the base to get the argument of the logarithm. So basically in this case x is essentially, in base 2, log(5).
@BZKnowHow
@BZKnowHow 13 дней назад
very informative and easy way of teaching
@user-xz4sl5pt2m
@user-xz4sl5pt2m 4 месяца назад
Aah my favourite maths algebra those days ❤🎉
@The.Bible.Community.
@The.Bible.Community. 2 месяца назад
How does this relate to real world practicality, thank you.
@user-ri8qh5tg1w
@user-ri8qh5tg1w 21 день назад
은가은씨 오늘도 즐거운 불금 되세요.....
@mouthiknaradas962
@mouthiknaradas962 2 месяца назад
Use substitution. Substitute 2^x =y and solve polynomial equation.
@MrIvanbrewer
@MrIvanbrewer 8 месяцев назад
reminder me my school years in one of the best math schools in Russia 25 years ago, now all forgotten but still these problems are solvable on the fly almost🙂 good times it was
@ArcobalenoOfLies
@ArcobalenoOfLies 8 месяцев назад
Какая школа?
@timothyfriedman8461
@timothyfriedman8461 10 месяцев назад
2^log²5 = 5 (by definition) 8^log²5 = (2^3)^log²5= (2^log²5)^3= 5^3=125 There is no need to do approximete calculations.
@andreadanieli6192
@andreadanieli6192 10 месяцев назад
In Italy this is called "col senno di poi"! 😂
@Psykolord1989
@Psykolord1989 Год назад
Before watching: This is not one that can be easily solved by simply plugging in integers and hoping for a result. Our solution is going to be somewhere between 2 (8^2 + 2^2 = 64+4=68) and 3 (8^3 + 2^3 = 512+8 = 520), and probably closer to 2 than to 3. Therefore, we have to actually do some calculations. Alright, 8 = 2^3, and (a^b)^c = a^(bc). Thus 8^x = (2^3)^x = 2^(3x). We declare U = 2^x. Then 8^x = 2^(3x) = u^3. Then we have u^3 + u = 130. Subtract 130 from both sides to get u^3+u-130=0. Now, we attempt to factor this. The factors of 130 are 1, 2, 5, 10, and 13. Of those, U=10 gives us results far too large, and U=2 gives us ones far too small. U=5 gives us 125+5-130 = 0, which is accurate. Thus, (u-5) is a factor. after doing some division, we can factor the equation into (u-5)(u^2 + 5u + 26 )=0. Going to use the quadratic formula on the second factor, we note that the discriminant is negative. Thus, these are not real roots, so we can skip this section. Thus, we will go with U=5. However, we're not done! That's the solution for U, not for X. U = 2^x. Thus, we have 2^x = 5. Take log_2 of both sides: Log_2(2^x) = log_2(5) -> X = log_2(5) Log_2 of 5 will be between log_2 of 4 (2) and log_2 of 8 (3), likely closer to 2 than 3. This checks out. If you want to change this to a different log using change of base, you may do so. log_b(a) = (log_c(a))/(log_c(b)),. Then using natural log ln, with a =5 and b = 2: x = (ln 5)/(ln2). (We're not doing this for the sake of precision, but rather so it can be easily checked on a calculator. A lot of calculators don't have functions built into them for logs of different bases, at least not ones that you can get to easily. Thus, you have to switch to either common log (log_10) or natural log (log_e))
@danypriandoyo6956
@danypriandoyo6956 8 месяцев назад
Superb
@forever_2791
@forever_2791 2 месяца назад
☠️
@kd8opi
@kd8opi 2 месяца назад
This is hard core algebra. Cubic equations, quadratic equations w/ the quadratic formula, logarithms, fractional exponents, ect…. I did stuff close to this level in high school. The difference was that it did not have multiple layers of this complexity. My takeaway from that experience was that algebra wasn’t hard so long as you worked a lot of different problems and got plenty of practice.
@abdellahaitouahmane1593
@abdellahaitouahmane1593 11 месяцев назад
To complicate solution We can divide by 2^x and y= 2^×
@user-ql9fu2dr1x
@user-ql9fu2dr1x 8 месяцев назад
..am I the only one to solve it in their mind after seeing the thumbnail?
@tapajyotipaul
@tapajyotipaul Год назад
Great session
@user-ym7lw7vd1w
@user-ym7lw7vd1w Месяц назад
if only the olympiad problems were this nice, solved in 2 seconds
@askthewise
@askthewise 7 месяцев назад
Was in it better to approximate x between 2 and 3 at a glance?
@fernandocarvalho2168
@fernandocarvalho2168 8 месяцев назад
Ótima explicação. Mas, quando você já havia encontrado que 2 elevado a x era igual a 5 , já poderia ter usado a definição de logaritmos e chegar direto na conclusão que x é igual a log de 5 na base 2 . Ou fazer assim seria um erro matemático? Parabéns pelo excelente vídeo!
@alexeyfadieiev4070
@alexeyfadieiev4070 8 месяцев назад
I also wondered why we need last manipulations with Log, it is redundant. 2^X=Y. X = logY.
@itsme.01
@itsme.01 8 месяцев назад
For 2^x=y y³+y=130 y(y²+1)=130 Clearly for y=5, equation satisfied so x=log5(base 2) is the correct answer.
@kevinsolari2744
@kevinsolari2744 8 месяцев назад
Having got to y³+ y = 130 it's not hard to try a few small numbers and see that y=5 is a solution. From that the quadratic part could be worked out, though it should be obvious that there can be no other real solutions since y>5 would be too big and y
@SanalDersaneLeventYadrga
@SanalDersaneLeventYadrga 7 месяцев назад
Yes I think so
@gorbachevaol
@gorbachevaol 7 месяцев назад
Два графика функций у=х^3 и у=-х+130 пересекаются в одной точке в 1 четверти=> х=5 единственное решение
@nasimthander9137
@nasimthander9137 8 месяцев назад
Your handwriting and logical thinking ability are awesome ❤
@mircorichter1375
@mircorichter1375 8 месяцев назад
I think the question is ill phrased because the types of the symbols in the equation are not specified. For example: no solution over integers. One solution over reals. 3 overvcomplex numbers. But what if we consider prime fields F_p with p>130 or any other algebra type where those symbols could be interpreted in?
@alexandrosmalamatis4746
@alexandrosmalamatis4746 8 месяцев назад
Beautiful!
@prashantprakhar3107
@prashantprakhar3107 5 месяцев назад
Your voice is so soothing
@math001
@math001 8 месяцев назад
Really crazy how easy this problem would be for my past self studying engineering. Being done with school and doing the same shit over and over again at work really rots your brain
@sergeykupcov4348
@sergeykupcov4348 5 месяцев назад
Это не легкая задача
@Peter-Alexander
@Peter-Alexander 4 месяца назад
Challenge your brain in your free time or find a more interesting job (when possible) 😊
@math001
@math001 4 месяца назад
@@Peter-Alexander yeah man I'm learning an additional 2 languages right now and also learning more music theory. Trading is also my side thing so I think the analytical part of my brain is still working to some extent. It's just that complex math isn't really my thing these days
@kolomun
@kolomun 5 месяцев назад
I think you could simply write 2^x = 5 => x = log2(5) without all these log divisions, because log2(5) literally means power to which we have to rase 2 in order to get 5, which is x in our case
@user-hn8uj5jo4m
@user-hn8uj5jo4m 8 месяцев назад
Спасибо! Не понимаю концовку с таймкода 4:38, и так ясно, что логорифм - это степень числа по основанию.🧐
@alokranjan4149
@alokranjan4149 6 месяцев назад
Beautiful question. It's answer is log5/log2 both on the base 10 👌👌
@jiminkook4445
@jiminkook4445 4 месяца назад
Anyone know which pen shes using? Looks so smooth 🥹
@carpacciodimanzOO
@carpacciodimanzOO 5 месяцев назад
how to complicate simple stuff
@XCodeHelpHub
@XCodeHelpHub Месяц назад
This is an imperfect solution. I realized that the power I'd 8 could not be 3 abs 2 was too small, meaning it was not going g to be an integer. Kind of ridiculous in my book.
@thedeathofbirth0763
@thedeathofbirth0763 4 месяца назад
To all the pretentious people who keep commenting that she made it hard, please note that she is trying to teach everyone at any level, that's why she chose the most straightforward method that any type of student can follow. She is not teaching only the undiscovered geniuses such as yourselves.
@BobHooker
@BobHooker 2 месяца назад
I love these, I do data analysis for word and they have no practical application but I still love them
@millaarsenal6291
@millaarsenal6291 4 месяца назад
Yeah I got it too. But when you check it, you should not convert to log value and you can simply find it.
@PaulJosephdeWerk
@PaulJosephdeWerk 11 месяцев назад
If you are asked for only the Real solution then youbare correct. If you want all solutions, you must take the quadratic into account and get two more Complex solutions.
@planomathandscience
@planomathandscience 9 месяцев назад
Don't be clever. When is a maths solution going to ask for complex solutions
@asliceofjackie91
@asliceofjackie91 8 месяцев назад
​@@planomathandscienceanywhere above high school, and at high school levels in some countries, expect a full answer unless otherwise specified.
@ab_random
@ab_random 8 месяцев назад
What happened in the end? Why the author used approximately solve? 2^log(2 5) = 5 by log's definition. And 8^log(2 5) = 2^3log(2 5) = (2^log(2 5))^3 = 5^3 = 125
@MathMathX
@MathMathX 3 месяца назад
Use rational zero theorem. Y=5. And divide the function by (y-5) and get the quadratic. Much faster....
@jurajchobot
@jurajchobot 4 месяца назад
Lol, I've done that independently from RU-vid thumbnail and my result was X = 1 / [ log130(10) ] which is X = log10(130) which reads as [ log 130 to base 10 ] Edit: It was quite close as my result evaluates to 2.1139 and the actual result evaluates to 2.3219
@DineshSingh-xv3bu
@DineshSingh-xv3bu 5 месяцев назад
Wawoooo. So simply explained
@henrypitch405
@henrypitch405 8 месяцев назад
Thank you 🥰🎉👏
@agilkes
@agilkes 8 месяцев назад
I have some concerns.......
@ankitasingh9884
@ankitasingh9884 8 месяцев назад
how can we split the middle term like this . u wrote 26 and 5 then how can you take 26 and 25 in tge next step
@AirCrou
@AirCrou 8 месяцев назад
Instead of that lengthy solution, let 2^x=y, and then let f(y)=y^3+y-130 For y=5, f(5)=0 f'(y)=3y^2+1, which is positive for all values of y, meaning f(y) is a monotonically increasing function, which makes y=5 the only root of f(y) Then 2^x=5 and solve using logarithmic properties
@StarLord1994
@StarLord1994 7 месяцев назад
Yeah, but this is only shorter because it happened to be monotonically increasing. If it wasn’t, you would’ve done this step for no reason, and still had to do the lengthy solution.
@b213videoz
@b213videoz 8 месяцев назад
y(y²+1) = 5(5²+1) ...at this stage it should be apparent that y = 5 🤪 Given y = 2^x Then 2^x = 5 log²(2^x) = log²(5) x = log²(5)
@minhsanghoangtran4653
@minhsanghoangtran4653 Месяц назад
This is a very easy question for high school students in Vietnam
@wisemang73
@wisemang73 Месяц назад
Nicely done. Weird how your 2's are written differently even in the same equation
@mku12
@mku12 8 месяцев назад
I think the easiest solution is that we know 2^7=128 next is 8^(1/3) = third root 8 =2 so 128+2=130
@ukaszgolanski8153
@ukaszgolanski8153 8 месяцев назад
This was my solution as well. Much simpler.
@prasoonrajsinghrathore7624
@prasoonrajsinghrathore7624 8 месяцев назад
Both the exponents are “x” so the value needs to be same. Can’t be 7 for one x and 1/3 for the other
@sleuthkonan
@sleuthkonan 12 дней назад
Yes 130=26x5, but how to get y=26y-25y ? I mean how to think of such an inference ?
@muntahajamil
@muntahajamil Год назад
Waiting for more videos, Mam
@naveenlakhara_51
@naveenlakhara_51 6 месяцев назад
Although I knew the answer to the problem, being lazy I preferred "Hit and Trial Method" from the given option for these type of questions. This saves a lot of time 😅😅
@ace_5639
@ace_5639 8 месяцев назад
Time to use synthetic division 😁
@kadir7533
@kadir7533 3 месяца назад
Its a mid level math problem for college exam in Turkey.
@dVTHoR
@dVTHoR 8 месяцев назад
I looked at this for like 2 minutes without a thought of any complex maths and thought the answer might be X= 2.25 and I’m honestly pretty pleased. Lol
@dVTHoR
@dVTHoR 8 месяцев назад
I also have zero training in mathematics outside of high school 10+ years ago so go very easy on me
@tijanimaths6006
@tijanimaths6006 8 месяцев назад
Thanks
@anilsengul1257
@anilsengul1257 4 месяца назад
Why cant you just add the left side to 10^x and do logarithm? Did I forgot of the basics of math? xD Please answere me I am really curious.
@CrunchurixASIS
@CrunchurixASIS 8 месяцев назад
I didn't get the last thing you did aa y-5= 0. If we can do this, then can't we do this for (a+2)(a-3)= 14 as a+2= 14 and a-3= 14? And that would give a completely wrong answer as the value of a should be 5. Please explain this.
@gopal-12
@gopal-12 8 месяцев назад
In the equation ((y - 5)(y^2 + 5y + 26) = 0), the Zero Product Property comes into play. This property states that if the product of two or more factors is zero, then at least one of those factors must be zero. This is why we can individually set each factor to zero to solve for the variable. So, for the equation:(y - 5 = 0) and (y^2 + 5y + 26 = 0) This process is different for (a+2)(a-3) = 14. In this case, the product wasn't set to zero, so we couldn't individually equate the factors to 14. The context there was different.
@JOpethNYC
@JOpethNYC 8 месяцев назад
I'm already confused at 1:29 . Where did the 25 come from?
@user-iu1yt6tz9u
@user-iu1yt6tz9u 5 месяцев назад
130 = 26х5
@hypocondria6835
@hypocondria6835 8 месяцев назад
Love this, Thanks ❤
@antoniomadrazo9919
@antoniomadrazo9919 8 месяцев назад
😊😊😊Vivian Lhu 0:14 0:14
@ankitsbizniz2010
@ankitsbizniz2010 3 месяца назад
Did this for a major part of life. Never used it again
@mediumQQ
@mediumQQ 8 месяцев назад
I have simplier solution. Multiply both by root x, so then 8+2 = x base root of 130. Which easily = 2.3....
@neilmccafferty5886
@neilmccafferty5886 5 месяцев назад
well explained.
@ycivdruhfvfjk5679
@ycivdruhfvfjk5679 6 месяцев назад
U can use to Ln function
@notsuraj_
@notsuraj_ Год назад
i solved it like this after 2^x(2^2x+1) = 130 so taking factor of 130 as two terms like 65X2 or 26X5 or 13X10 and trynna see which of these two terms satisfy that eq which is on taking 2^x= 5 so thats the answer after taking log of this
@antoniou.1158
@antoniou.1158 8 месяцев назад
Nice
@mocotone
@mocotone 8 месяцев назад
As soon as she got it to Y Cubed plus Y = 130, the answer jumped out at me and I yelled it out loud, trying to beat her to the punch. Who knew I actuality had time to go check the mail, first?
@free11192
@free11192 9 месяцев назад
Log 2^x = x log 2. Please learn this. For your solution: X= log5/log2.
@BERNDWERK
@BERNDWERK 8 месяцев назад
Nice, but you should see, that y^2+5y+26=0 can't have any real solutions without computing by seeing, that it's graph the shifted y^2-graph into positive direction.
@surajitde8537
@surajitde8537 9 месяцев назад
This can be solved by vanishing factor method By putting y=5 y-5 is a factor 5^3+5-130=0 y^2(y-5)+5y(y-5)+26(y-5)=0 (y-5)(y^2+5y+26)=0 y-5=0 or y^2+5y+26=0 y=5. D for quadratic equation 5^2-4×1×26 is less than 0 No real roots. Solution will be y=5
@mairepcod4063
@mairepcod4063 8 месяцев назад
Thanks,
@glorypath810
@glorypath810 8 месяцев назад
Go easier by using Factorizing with polynomial and basic logarithm
@juneldomingo6277
@juneldomingo6277 8 месяцев назад
In real exams, you can just substitute the given choices if satisfies the 130. It will save lot of time instead of solving.
@AceGunner72
@AceGunner72 8 месяцев назад
In real exams you are not given choices. What educational system did you attend? "Fast food and exams Inc."?
@juneldomingo6277
@juneldomingo6277 8 месяцев назад
@@AceGunner72 maybe you didnt take any licensure exams.
@alfredchan590
@alfredchan590 6 месяцев назад
I hate log, because every time you calculate the log, it is impossible to calculate by yourself without calculator.
@EnochAsamoah120
@EnochAsamoah120 5 месяцев назад
Please I don't understand why you multiplied the 5y by 5 to insert into the equation
@Yes_I_c4n
@Yes_I_c4n 11 месяцев назад
Why aren't the imaginary solutions taken into account? I mean, they are solutions to the original equation, aren't they? Is there an assumption that we must find the real solutions only?
@bishtkunal10
@bishtkunal10 5 месяцев назад
I literally solved it in my head, this is too easy to be an olympiad question
@god_bika
@god_bika 8 месяцев назад
4:47 x is already log 5 to the base 2 as per logarithm definition,. Isnt it
@sujeetiitd
@sujeetiitd 5 месяцев назад
We can do it faster. 130 has only 3 non repeating prime factors 5, 13 and 2. So there is not much of a substitution job to check which one satisfies Y * (Y^2 +1) = 130 And its 5.
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