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

@nitdiver5

3 ай бұрын

Or funny cats

@Aman-nk5uq

3 ай бұрын

Come on. Everyone should know how to solve this

@xtranub8792

2 ай бұрын

​@@Aman-nk5uq he means it's too easy

@Fingolfin3423

Ай бұрын

Most mathematicians probably can't cook worth a damn. Most people can't.

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

7 ай бұрын

(8×15)+(2×5)=130

@noobLOL77

7 ай бұрын

thats what i did, i got hung up because i did log(5)-log(2) instead of log2(5)

@IvyANguyen

7 ай бұрын

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

7 ай бұрын

y=2x , find the value idiot

@DantesInferno61

7 ай бұрын

She made it way harder to solve than she had to.

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

8 ай бұрын

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

7 ай бұрын

@@tintiniitkwe learned it as logba=e where e is the exponent b is the base and a the “answer”. its very helpful

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

7 ай бұрын

Wowwww😮😮😮

@July-gj1st

7 ай бұрын

Nice! I did it like this as well or, similar.

@StarLord1994

7 ай бұрын

Yeah, it just depends. A lot of times that advice will work, but not always.

@zhenyuzhai4098

2 ай бұрын

Yours can't guarantee there is unique root.

@georiashang1120

2 ай бұрын

@@zhenyuzhai4098 I know it should be unique in this case

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

7 ай бұрын

Looks like you're stuck in a loop in life.

@krishhhhhhhhhhhhhhhh

7 ай бұрын

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

7 ай бұрын

*for 15 years

@ankurmondal3220

7 ай бұрын

​@@krishhhhhhhhhhhhhhhhthese type of outside the box question are what asked in sof olympiads

@Smart_Soham

7 ай бұрын

​@@ankurmondal3220so you are studying fkr sof olympiads, go and give the real IMO Make sure you don't piss in your pants

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

@RedPardo

8 ай бұрын

Came here to note that.

@manojpadmanabhan2615

7 ай бұрын

Exactly.. 2^(log 125/log 2) = 125 and so on.

@antoniou.1158

7 ай бұрын

​@@manojpadmanabhan2615 let's be nice with our math mates

All these problems seem a little too easy for olympiads

@estefanocrespo7930

7 ай бұрын

Are you sure?

@human18711

7 ай бұрын

​​@@estefanocrespo7930yes these are like middle school problems for us

@ritsuboii2130

7 ай бұрын

​@@estefanocrespo7930im not him but yeah this is pretty easy for olympiads because i can even answer it

@alexcwagner

7 ай бұрын

Maybe it's the special olypiad

@gauravbasu98

7 ай бұрын

😂😂😂😂

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.

The definition of the logarithm is "to what power do I need to raise value1 to get value2. So you could write from start that x = log(2, 5).

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

7 ай бұрын

I agree, the finish could have been done a bit more elegantly with out finding the decimal approx.

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

7 ай бұрын

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

7 ай бұрын

Same thoughts here. I literally solved it in mind using the similar way as you in 1min

@m3zuss

7 ай бұрын

Yeah but try to solve it when u were 13 or 15.

@Quiablo

7 ай бұрын

Damn we really dont need a 8 minute to solve this one. Nice observation.

@shashankkhatri5523

7 ай бұрын

people @13-15 in india can solve this way faster compared to the older folks in india btw@@m3zuss

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

7 ай бұрын

It’s just to prove that there is no other solutions with real values

@daakudaddy5453

7 ай бұрын

You also have to check for other possible solutions. Duh!

@adriendebosse6941

7 ай бұрын

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

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

9 ай бұрын

The imaginary solutions were extraneous. That's why they were explicitly discarded.

@rickymort135

7 ай бұрын

​​@@NicholasOfAutrecourtextraneous or extra anus?

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 .

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.

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.

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.

This is why I know anatomy so well!

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

7 ай бұрын

I also wondered why we need last manipulations with Log, it is redundant. 2^X=Y. X = logY.

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

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.

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

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

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).

Your handwriting and logical thinking ability are awesome ❤

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).

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

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

8 ай бұрын

Yeah, exactly my thoughts: just use the definition!

@intigamaghamuradov4727

7 ай бұрын

It was painful to see how she derives x after getting 2^x = 5 (((

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

7 ай бұрын

Yes I think so

@gorbachevaol

7 ай бұрын

Два графика функций у=х^3 и у=-х+130 пересекаются в одной точке в 1 четверти=> х=5 единственное решение

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

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.

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

7 ай бұрын

Какая школа?

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.

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

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

9 ай бұрын

In Italy this is called "col senno di poi"! 😂

The longest way to do that equation.

Aah my favourite maths algebra those days ❤🎉

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

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

4 ай бұрын

Это не легкая задача

@Peter-Alexander

3 ай бұрын

Challenge your brain in your free time or find a more interesting job (when possible) 😊

@math001

3 ай бұрын

@@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

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

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 !

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.

@planomathandscience

8 ай бұрын

Don't be clever. When is a maths solution going to ask for complex solutions

@asliceofjackie91

7 ай бұрын

​@@planomathandscienceanywhere above high school, and at high school levels in some countries, expect a full answer unless otherwise specified.

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?

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)

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

7 ай бұрын

Superb

@forever_2791

Ай бұрын

☠️

Anyone know which pen shes using? Looks so smooth 🥹

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

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

7 ай бұрын

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.

Your voice is so soothing

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

Beautiful!

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 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

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

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

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

Wawoooo. So simply explained

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

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

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

Thank you 🥰🎉👏

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

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

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

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 😅😅

Waiting for more videos, Mam

Go easier by using Factorizing with polynomial and basic logarithm

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

7 ай бұрын

I also have zero training in mathematics outside of high school 10+ years ago so go very easy on me

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

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.

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 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

I have some concerns.......

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

@user-iu1yt6tz9u

4 ай бұрын

130 = 26х5

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

7 ай бұрын

Nice

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.

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!😊

Love this, Thanks ❤

@antoniomadrazo9919

7 ай бұрын

😊😊😊Vivian Lhu 0:14 0:14

well explained.

U can use to Ln function

What grade of math Olympiad is this for?

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

I could have just put some value in place and solved mathematically to make a guess that i have to go up or down a little. Than three attempts later i might have got x=2.3 I have done it many times. You dont need to go in multiple algebrac complications to solve a problem this simple. Afterall numbers are used to avoid hit and trial of actual objects. We can use numbers for hit and trial as that wont even use extra ink or time cz the equations are more complex than hit and trial. Smart work is also intelligence!

Let y=2^x then y³+y-130=0 Easy to see y=5 is one solution. I will skip complex solutions which just requires synthetic division and quadratic formula to solve. 5=2^x -> log_2(5)= x -> x= ln5/ln2

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

7 ай бұрын

This was my solution as well. Much simpler.

@prasoonrajsinghrathore7624

7 ай бұрын

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

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?

Can someone show how to solve it with a graph?

I learn my level is olimpic thanks to this channel.😂

How 25 came during factorisation?

Do you have a short cut to solve this....back to 1994 my tutor can sovle this in 2 minutes

this is extremely easy for an olympiad question

Thanks

It's not quadrilateral equation so how can u factorize Middle term like this

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

@AceGunner72

7 ай бұрын

In real exams you are not given choices. What educational system did you attend? "Fast food and exams Inc."?

@juneldomingo6277

7 ай бұрын

@@AceGunner72 maybe you didnt take any licensure exams.

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

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

Can't we just use log on both sides?

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

@user-iu1yt6tz9u

4 ай бұрын

У = 26у - 25у У = 1у