Mexico - A Nice Math Olympiad Exponential Problem 

This is why I'm going back to watching cooking videos.

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.

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.

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)

All these problems seem a little too easy for olympiads

Instead of using decimals you could have used more logarithmic identities when you were checking. I think that would have been cleaner.

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.

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.

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

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.

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 😅

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.

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.

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

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.

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 .

Confuse the number 26y - 25y, can you explaine it!

This is why I know anatomy so well!

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.

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!

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

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.

Thanks you so much for refreshing my memory from 10 years ago, Now I wish I have continued on the math field instead

The longest way to do that equation.

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.

So hard. Congrats. Several math concepts 👏👏👏👏👏

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

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).

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

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).

Спасибо! Не понимаю концовку с таймкода 4:38, и так ясно, что логорифм - это степень числа по основанию.🧐

Your handwriting and logical thinking ability are awesome ❤

very informative and easy way of teaching

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?

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

Aah my favourite maths algebra those days ❤🎉

2^log²5 = 5 (by definition) 8^log²5 = (2^3)^log²5= (2^log²5)^3= 5^3=125 There is no need to do approximete calculations.

To complicate solution We can divide by 2^x and y= 2^×

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.

how to complicate simple stuff

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.

How does this relate to real world practicality, thank you.

I think 5 is a rather easy to find "obvious" solution by searching a integer which cube is close to 130. Then it becomes a polynom division. The proposed factorisation is nevertheless very smart !

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?

Was in it better to approximate x between 2 and 3 at a glance?

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))

Beautiful question. It's answer is log5/log2 both on the base 10 👌👌

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

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

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.

Great session

..am I the only one to solve it in their mind after seeing the thumbnail?

Use substitution. Substitute 2^x =y and solve polynomial equation.

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.

if only the olympiad problems were this nice, solved in 2 seconds

I have simplier solution. Multiply both by root x, so then 8+2 = x base root of 130. Which easily = 2.3....

Use rational zero theorem. Y=5. And divide the function by (y-5) and get the quadratic. Much faster....

Yeah I got it too. But when you check it, you should not convert to log value and you can simply find it.

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

Factors of -130 according to you are 26 and -5. Why did you use 26 and-25.?

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

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)

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

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 😅😅

great video but the verification can be a little bit better if it's like this: 8^log5 base 2 + 2^log5 base 2 = (2^3) ^log5 base 2 + 2^log5 base 2 = (2^log5 base 2) ^3 + 2^log5 base 2 as we already know a ^logN base a = N => (2^log5 base 2) ^3 + 2^log5 base 2 = 5 ^3 + 5 = 125 + 5 = 130 excellent video keep it up and upload more videos like this.

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.

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

I love these, I do data analysis for word and they have no practical application but I still love them

Nicely done. Weird how your 2's are written differently even in the same equation

In real exams, you can just substitute the given choices if satisfies the 130. It will save lot of time instead of solving.

You make the solution to obtain the real part seem too complicated. from the point where (2^x)=5 by just using the conversion law from Exponents to log we obtain Log5_2 = x. Even to check the result, using logs instead the decimal produces a more exact result or (125+5=130).

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

Time to use synthetic division 😁

Beautiful!

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.

I have some concerns.......

Waiting for more videos, Mam

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

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

This is a very easy question for high school students in Vietnam

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?

I think it's not of Olympiad standard . Btw I am Indian , so we often do these problems in school . ( Please don't get offended)

Its a mid level math problem for college exam in Turkey.

Anyone know which pen shes using? Looks so smooth 🥹

I got from a much easier and faster way 2*3x and 2*x should be equal to 130 so the sum of two of 2 power something should be equal to 130 which is close to 2*7 hence if x=2 the answer is 64+4 and if x= 3 it would be 528, therefore, x should be between 2 and 3 and closer to 2 than 3 and 3x kinda feels be close to 7 as 128 is close to 130 so 7/3=2.33 and as it wouldn't add up (cuz 127+4,9 is greater than 130 we should decrease 2,33 by 0,01 per time to find the right answer then we got the answer of 2.32

If we simply check in starting of cubic equation that one of its root is 5 as 5^3+5-130 equals zero, it can me much easliy solved

Go easier by using Factorizing with polynomial and basic logarithm

은가은씨 오늘도 즐거운 불금 되세요.....

Your voice is so soothing

Thank you 🥰🎉👏

I literally solved it in my head, this is too easy to be an olympiad question

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.

Please I don't understand why you multiplied the 5y by 5 to insert into the equation

Gives me shivers, brings back nightmarish memories of struggling with all this in school years!😮 But enjoying it too, in a contradictory way, because there is no pressure!😊

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.

Thanks

The logarithm part could've been done quicker, but I see a lot of comments saying to just observe that y^3+y=130 means y=5. Sure, you are not wrong, but you skip checking if other real solutions exist, thus giving a potentially incomplete answer.

I'm already confused at 1:29 . Where did the 25 come from?

Log 2^x = x log 2. Please learn this. For your solution: X= log5/log2.

1:37 is confusing? Why and how can you write y=26y-5y?