@@beaumatthews6411 This question was undoubtedly first written long before calculators were invented, hence the simple solution x=log(5)/log(2) The answer would never be expected in log base 2 because the log tables used before calculators only had log base 10 (log) and natural log (ln). The final answer would be obtained using a slide rule. Fun methods that put a man on the Moon.
Once we had y^3 + y - 130 = 0, it was immediately clear that y = 5 and the problem was solved. I’m not sure that the approach of changing y to 26y - 25y is an easier approach; it seems it requires more insight. Your thoughts please.
You divide the polynomial with y-5 and check the remainder, which is y²+5y+126. It's clearly a polynomial division issue. You should check the roots of the remainder because they could be real numbers too. So in general a solution other than 5 could be also feasible.
too long , when you let k = 2^x, let f(k) k^3+k-130=0 just find the factor of f(k), that's 5 ,(when k=5,f(k)=0),so that k-5 is the factor of f(k), use the long division method, we get (k-5)(k^2+5k+26)=0 k=5,k= (-5+/- sqrt(79)i)/2 2^x =5 x=log(2.,5) x=2.32 (3 con fig)
@@anson2075same thing really, cardano is more general to apply to other problem. Though a faster is observe x=5 is a solution of x^3 + x = 130, because x^3 + x is strictly monotonic so x=5 is the only real solution
I just brute forced this knowing that 8*8 is 64 and 8*8*8 is 512 it would have to be between 2 & 3 then via estimation and guess and check 2.32 came the closest in approximation. Even though I know this method is wrong on a standardized test with multiple questions and limited time. It is effective having a “feel” for numbers and proportions gets you through most of these tests
Used same method in about 2 minutes. Might have been able to do it analytical 40 years ago.. Hardly used any analytical math since poly technical. It's now a distant memory.
From the line y^3+y-130=0 (y^3-125)+(y-5)=0 (y-5)(y^2+5y+25)+(y-5)=0 (y-5)(y^2+5y+26)=0 This way it took me 3 extra lines to get to the same result that the admin here took 6 lines.
At the second step, if you can see that 2^x = 5, you have the answer, and you'll have time to answer some other questions. These questions usually rely on having an insight like this, rather than on grinding out a solution, so you should be asking at this point, "is there a cube slightly less than 130?"
A guess is a guess. Without demonstrating thinking process, replacing y = 26y - 25y is nothing but a guess, which teaches no general approach to solving similar problems. It is no better that noticing that y^3 + y grows pretty fast, so we only need to try a few integers to get 130. And 5 pops up almost instantly. It’s also a guess, but much more natural and less contrived.
agree with you, applying y = 25y - 25 y is far from a demonstration process, it is simply a trick. I prefer to use my brain and notice, as you did, that y^3+y grows fast
Isn't that what mid term split is all about tho? We deliberately choose those factors that gives the constant on multiplying and the coefficient of the lesser degree variable in the equation on adding?
equations can have irrational, complex and non integer roots, which are often impossible to guess (although they can be approximated). What you made was a guess for an integer root, and its standard practice to check a cubic for an integer root first, and then use factor thereom. Sometimes, an integer root does not exist or is too large and you must use cleverer methods. You often need to find the complex roots (which were skipped over by the poster) It is not contrived to guess a particular factorisation. A lot of good algebra problems are about making clever observations, factorizations, splitting etc. Many good problems asked in olympiade and competetive exams all around the world are based on using clever little observations, which instantly simplify the problem. There are 'non contrived' ways to solve cubics, but they are absurdly long (check out the cubic formula. The derivation is pretty interesting). And they do not even exist for a polynomial equation greater than 4 degree.
Remember this being taught in school, when I was about 13-14 years old. Never once used it. Forgot it completely. Then, at age 43, I needed to pass an exam for a certain university and I was taught such things again. Never ONCE used it. Forgot it completely. This is like learning a language, that only few people speak for real in the world. I was taught several languages in school as well. Didn't know it at the time, but that has proven to be excellent usable knowledge, that I literally use each and every day in my normal life. The history lessons keep from falling for just any propaganda. The geography lessons gave me basic understanding of who's where in the world, science and biology brought some insight too. But THIS? Completely superfluous. An utter waste of time. This ought not to be taught in schools.
@@orlevene9964 Not normally left like that. The day will come when you have to use a calculator without a base 2 log function, so it is best to know how to change to base 10 or natural log.
As soon as I saw that we had a cubic to solve, and 130 has a prime factorisation to 3 primes, 2x5x13, I knew one of them was going to be a candidate! Also if you dont do the 26-25 trick, and consider what the coeficients of general cubic would be, given there is no y^2 coeficient, you get the property that the sum of the roots is 0, and the product of the roots is 26.
My approach before Watching this video put 2^x=t Now, we have a simple cubic equation t³+t=130 =>t(t²+1)=130 Now,by trail and error method Factors of 130 = 2×13×5 Putting t=5 in LHS,we get 5(25+1)=5×26=130 Hence the value of t=5 As,2^x=t (Taking logarithm both sides) =>xln2=lnt Put the value of t=5 x=ln5/ln2 Got the answer by simple algebra manipulation and took 2 minutes to write and seconds to solve
Once you have the cubic in t, just try looking for a perfect cube that's not much less than 130. Trying 125 = 5³, we have the solution, because 5³ + 5 = 130. The rest follows as you describe. Fred
Yup you just need basic 10th grade mathematics fundamental concepts and you can solve it easily within 3 minutes This made me realise how hard Jee Advance is.
I'm impressed to see so many alternative solutions. I see that the followers of this channel are true mathematicians. Well done! I would have solved the cubic equation using synthetic division. Best regards from Bogota.
I'm starting to freak out that after reading most of the comments I can't find any criticism of the line at 3:43 that claims y^3 + 25y = y (y^2 - 5) What am I missing here?
Using bracketing (artillery method for finding range of target) and trial and error method should be able to do it in one or two minutes which would be much faster and more importantly less prone to algebraic error, but of course less elegant. Also if it is a multiple choice test you could quickly try the different answers and find the correct one in very little time.
Let us replace 2^x by X, then X^3 + X = 130, and evidently X = 5 is a solution. As the function X -> X^3 + X is strictly increasing on R+, the value 130 is reached by the single X = 5. So the unique real solution is given by 2^x = 5, i.e. x = Log 5 / Log 2. Thank you for your videos !
8^x + 2^x = 130. 8 = 2^3 so we have 2^3x + 2^x = 130. You can say, immediately, that by inspection 2^x = 5 because 5^3 + 5 = 125 + 5 =130. Thus x = log5/log2 . I can't do that in my head - I know it's 2 point something and my best guess would be around 2.3 - but a few seconds on a calculator tells me that x = 2.3219 to 5 sig fig. Or if you need a little more time you can try 2^x = y so y^3 + y = 130 and again you can immediately see that y = 5. The rest follows - no need for all that manipulation for a simple problem which is what, I think, the Cambridge examiners would want to see. No need to use a sledgehammer to crack a peanut.
Interesting video, but if you're going to write the product as 26*5, why not just write it as 25*5 + 5 since you have y^3 + y, so 125+5 = y^3 + y, 5^3 + 5 = y^3 + y therefore y = 5 x = log_2(5).
8^x + 2^x = 130 (2^3)^x + 2^x = 130 (2^x)^3 + 2^x = 130 u = 2^x u^3 + u = 130 u^3 + u - 130 = 0 u=5 is a root (u^3 = 125, 125 + 5 - 130 = 0). So, 5 = 2^x log(5) = log(2^x) log(5) = x*log(2) x = log(5)/log(2) This does indeed solve the equation. There would be another pair of solutions, which we could get by factoring x-5 out of the cubic and solving by usual quadratic methods.
Lets write the complex roots, as well. y²+5y+26=0 ⇒ y=√26·[(-5±i√79)/(2√26)]. To evaluate log of y, we choose the branch cut of the log to be the non-positive real half-line & restrict the arguments of the complex numbers to lie between -π & π. Then, log₂(y)=log₂(√26·[(-5±i√79)/(2√26)])=log₂(√26·e^(±iθ)) for θ=cos⁻¹(-5/(2√26))≅.663π ⇒ log₂(y)=log₂(26)/2±iθ/ln(2) where "ln" is the natural log.
ln5/ln2 is the ans. I am currently in 11th and i did this by basic algebra. Assume 2^x = t and then youll get a cubic eqⁿ. One of its roots by inspection is 5, and the other remaining Quad. Eqⁿ is a always +ve factor, so it doesnt matter. Hence there is only 1 solⁿ, y=5 => 2^x=5, ln(2^x)= ln5, x(ln2)=ln5 x=> ln5/ln2 or log5 base 2👍🏻
Algorithm gave me this after looking up beginner algebra for my 7 year old (she wanted to know!). I haven’t done this math in, oh, 10 years and some of it before then. So while everyone else can spot the obvious based on the comments, I really did enjoy the extra explanations. I could follow fairly well despite not having used those neural pathways in a very long time. Perhaps if someone had done so when I first leaned this math, I’d be an MD instead of an attorney!
Oh okay I see you went for the longer method. The moment I saw the question the intuition was very clear because of exponential I immediately used log on both sides, and put the values to the calculator.
The answer to the title card is: (8 x 13)+(2 x 13)=130 X=13 Technically it is simple division but you get there by multiplication. You take the 130 as your cap and see how many times you can multiply the 8 to fit in there. While leaving room for the 2 times whatever, by repeating the process for the 2 and adjusting the 8's multiple as needed you'll get to the point the that you're multiplying both numbers by the same ammount to fill out the cap. Which is what you want in this case. 8x + 2x = 130 ( 8 × *X* ) + ( 2 × *X* ) = 130 ( 8 ×13 ) + ( 2 × 13 ) = 130 8 × 13 = 104 2 x 13 = 26 104 + 26 = 130 X = 13
At 2:12 you missed a huge opportunity to use the Rational Root Theorem. If you had used it, you could have shown right away that y=5 was a root. You could then have used polynomial division to find the other factor as y^2 + 5y + 26. Instead you applied 26y - 25y as magic without any justification. This does not help the student.
If you like base-10 logs, log(5) = log(10/2) = log(10) - log(2) = 1-log(2), so you get x = 1/log(2) - 1. We know log(2) = 0.30103, so we can quickly compute it.
7:40 Since you've mentioned complex solutions, 2^x=5 has a foliating solution in ℂ, just like any logarithm. By the fundamental theorem of algebra, there are 3 solutions in ℂ to the cubic polynomial in y, so there were not even a need to compute D to find out that there are 2 more solutions for y in ℂ\ℝ. That there is only one y∈ℝ⁺ is clear if you rewrite y³=130−y and consider both sides as functions ℝ⁺→ℝ⁺: [y³=f(y)]=[g(y)=130−y]. Both f and g are continuous and surjective, so f=g has at most one solution. Note that f(y) is also monotonically increasing, g(y) decreasing in y, so there exists a solution. The other two got to be complex. Just for the fun of it, since 2^x=exp(x log 2), the solution to exp(x log 2)=5 is foliating in ℂ: x=(log 5/log 2) + 2πki/log 2, k∈ℤ. x∈ℝ only for k=0, obviously. The complex roots of y²+5y+25=0 are not only extremely ugly, as you correctly noted, but also ought to foliate with the same period 2πki/log 2.
The first thought is to replace 8 with 2^3, then cubic equation follows. We do not expect kids to remember Tartaglias formula so this is a dead end. If we hope for a integer solution it is good to remember that the free term (130) is a multiplicative of all 3 roots. Indeed it can be factired as 2*5*13. By checking al divisors of 130 (2,5,10, 13, 26, 65) it turns out that 5 is a solution and it is easy to see that it is also the only real solution.
If you use prime numbers, the equation is easy to solve. That is 2^3x +2^x=(2)(5)(13) but this can also be written (2^x)(2^2x + 1)=(5)(26). And therefore 2^x=5 and x=log2(5).
Being a middle school student i thought that this is an algebra equation in one variable😂😂. I thought it is 8x + 2x = 130 and here x = 13 i was like it is so damn easy why are they freaking out😂😂😂
u³+u = 130. Think of a number whose cube + the number = 130. Obviously 125+5=130, so 2^x = 5. Hence x = log5/log2 = log₂5. Very easy. Did it in my head from the thumbnail!
Nice problem. It would be more helpful if you explained why the substitution y = 26y - 25y would actually occur to you at the step you take it, rather than just show that "it works out nicely". Why not instead notice that 130 = 13 x 10, and proceed with y = 13y - 12y? I.e., explain that because 25 is a perfect square you'll get a factoring present in a moment.
I gave this problem to the new ChatGPT model (o1) and it got the same answer and solved it in a very similar way (although without the 26y - 25y stuff). Impressed!
Well your overqualified for your job then and should look for a job that requires more intellect than tightening bolts because only 1% of adults in the general population could solve this problem.
y=5 because life's short and when faced with y^3+y=130 you check what you can before being forced to argue as you must. Are there other relevant solutions to the cubic? Picture the graphs of 2^x and 8^x for real x. On the LHS of the origin we have a small sum,
@@piotrstrzelczyk5013 same but i keep getting 2. I changed 130 to 2⁷ + 2. Changed the other side to powers of 2 as well Took log, then cancelled And ended up with 4x = 8
When I got my first programmable calculator, an HP25, I wrote a simple bisection routine to solve the countless equations I came up with when taking physics exams. That is still how I would choose to solve problems like this. I leave math to the mathematicians. Professors, who had been raised at a time when 4k of memory had to be rented because it was too expensive to buy, would tell me my approach was too expensive. I think about those comments every time I look at my stash of gigahertz laptops that are collecting dust.
Interesting. The solution hinges on writing 130 = 5 x 26, which is noted and exploited in the on-screen "working out". Similarly observed: 26 = 25 + 1 = 5^2 + 1. Hence 130 = 5 x (5^2 + 1) = 5^3 + 5. Comparing 5^3 + 5 = 130 with 8^x + 2^x = 130 suggests 2^x = 5 and 8^x = 5^3, the latter of which can be rewritten 8^x = (2^3)^x = (2^x)^3 = 5^3, confirming that 2^x = 5. Finally, 2^x = 5 => x = log5/log2, as stated on-screen.
I used to be amazed by calculus, I was never fluent, I was really bad memorizing all these little rules. This feels like someone playing guitar really well in front of me but I'm into electronic... Like dude I wish I had the same passion for this subject.
Before watching (so I might be very ashamed after!)... but it seems too easy. The BIG clue is that 8 is a power of 2 (2 cubed). So 8^x is (2^3)^x. And then we need a bit of substitution, so y for 2^x. Which means we have y^3 + y = 130 and it's clear there are very few options (as cubes get big quickly, so 6 is already a bust). It (y) must be 5. so 2^x = 5 and 8^x = 125. x is a bit more than 2 (2.2 or 2.3 ish). so i need a calculator to get exact number. or just express as logs as 2^x = 5 is same as x log 2 = log 5 therefore x = log 5/log 2. which is 2.322. And, of course, 8^2.322 + 2^2.322 is roughly 130. (130.019)
This was so damn easy like I had started preparing more jee about 4 months ago, and still solved it, like this exam zeems so easy and nothing compared to jee if these type of questions come in it
I make this: 8^x+2^x=130 (2^x)^3+2^x=5^3+5 If we named 2^x as Y. Therefore, we have: Y^3+y=5^3+5 Which immediately means that Y=5 So, 2^x=5. The equation show us that the answer is log 5 to the base of 2=x 😁
I’ll be honest with you I might be screwed I’ve done so many questions like these my first thought was “I’ll take the natural log of both sides” and my second thought was “maybe we can write in u as 3x my brain is so calculusified I can’t solve basic algebra
After substituting y, we could factor out y and find the product of 130. y(y^2+1)=130 y(y^2+1)=26×5 y=5 y^2+1=26 y^2=26-1 y^2=25 y=+5,-5 but y=5 is the same as th other one. Therefore since both y are equal then y=5 Then we can solve for x
But you cant just turn 130 into any 2 numbers you want to then set y equal to them multiplied together because there are infinitely many combinations. You could make it 2*65 if you wanted and get something completely different.
математикам известна одна единственная формула пифагора, и как только они ее не выкручивают чтобы получить целые области для одурачивания публики, при этом используют метод обратного реинженеринга чтобы прийти к заранее известному результату
8 EXP X = 4*2 exp x Re-writing the equa. 4*2 exp x + 2 exp x = 5* 2 exp x = 130; 2 exp x = 130/5; x log 2 = 26; x = log 26/log 2. Solve from there. Basically, 2 exp 6 = 128, so we're looking for x
I think your approach is more complicated and time-consuming (since it's an entrance exam and the time is limited, we should consider the easiest and the fastest approach to minimize the probability of mistakes and spend the least amount of your time). So, for me, it was faster and simpler to make in this way: 8^x + 2^x = 130 2^3x + 2^x =130 2^x = p; x=log p [base2] 2^3log p [base 2] + 2^log p [base2] = 130 p³ + p = 130 p(p² + 1) =130 Now we can just pick a number, the calculations are not complicated. We need to find a number that, when squared, is not too big (because it will then multiply by itself). We can consider the cubes of numbers from 3 to 6 because they are quite easy to calculate in your head. 3³=27, 4³= 64, 5³= 125, 6³=216. The cube of 5 is closest to 130, so we can easily and quickly check that 5 is the answer for p(p²+1)=130 (I think it is not difficult to calculate 26*5 in your head). If you have had experience with quick mental arithmetic at school or with your tutor, then to pick up number in this equation will take you no more than a minute and a half. p=5 x=log5 [base2]. Your thoughts
It is immedistely obvious that there exists exactly one real solution (the LHS is a monotonically increasing bijective function from the reals to the positive reals). Since 8=2^3 and 130=5^3+5 we have x=ld(5), where ld denotes the logarithmus dualis.
When you have 8^x + 2^x you have something cubed + something. That makes 130 = 125 + 5, so the something is clearly 5. Then 2^x = 5 and x = log5 / log2. Exponentials like these don't have multiple real solutions, because they are positive and monotonically increasing, so that's the only real solution. That takes less than 30 seconds, not 10 minutes, and you really need not to be hanging about on a Cambridge entrance exam.
Who says that you are interested in only real number roots? It does NOT state that in the instructions (which actually do not exist anyway), and you never stated it at the beginning of the video.
I’m just doing basic algebra with my son for his year 7 homework. Why isn’t it simplified to 10x = 130? So x = 13? That’s what my son has just learnt to do.
First thought: "Oh, that's simple, just 128 + 2, which is 2⁷ + 2¹" Then after diving in: "Something between 2 and 3, if they chose =68 or =520 this would be simple"
Is there some hidden agenda in writing x like an egypt hyroghlyp inthead of just to lines crossing ? Ah I got it it takes longer so you have more time to speak during writing. ;-)
2::19 he says: "130 can be read as the product of 26 times 5" --- 130 = 26 x 5 --- WOW! UNBELIEVABLE! --- For a) 130 = 65 x 2 is the same as b) 130 = 13 x 10 *WHY* did he not choose a) y = 65y - 64y or b) y = 13y - 12y ???
From 2^x = 5, x = log 5 to the base 2 directly follows. No need to do any algebraic manipulation.
Yes I also did same
He uses the theorem to prove the definition of logarithm.
For some reason they dothis shit all the time in school
@@beaumatthews6411 This question was undoubtedly first written long before calculators were invented, hence the simple solution x=log(5)/log(2)
The answer would never be expected in log base 2 because the log tables used before calculators only had log base 10 (log) and natural log (ln). The final answer would be obtained using a slide rule. Fun methods that put a man on the Moon.
I was going to say that. Cheers!
Once we had y^3 + y - 130 = 0, it was immediately clear that y = 5 and the problem was solved. I’m not sure that the approach of changing y to 26y - 25y is an easier approach; it seems it requires more insight. Your thoughts please.
There are two more solutions, though complex. I agree that if you can see the (y-5) factor, it’s probably faster to just do the poly division.
You divide the polynomial with y-5 and check the remainder, which is y²+5y+126. It's clearly a polynomial division issue. You should check the roots of the remainder because they could be real numbers too. So in general a solution other than 5 could be also feasible.
Ergo: x = ln(5)/ln(2)
@@jjeanniton why using a base other than 2? It is clearly a base 2 logarithm: log2(5)
Yep
With Cardano/Tartaglia formula we get the following reasonning:
8^x + 2^x = 130
(2^3)^x + 2^x = 130
(2^x)^3 + 2^x = 130
let k = 2^x
k^3 + k = 130
k^3 + k - 130 = 0
k^3 + k - 130 = 0 is based on k³ + kp + q = 0 template, with:
• p = 1
• q = -130
recall and application of the Cardano/Tartaglia formula:
• k = [-q/2 + √(q²/4 + p³/27)]^(1/3) + [-q/2 - √(q²/4 + p³/27)]^(1/3)
k = (-(-130)/2 + √(((-130)^2)/4 + ((1)^3)/27))^(1/3) +
(-(-130)/2 - √(((-130)^2)/4 + ((1)^3)/27))^(1/3)
k = 5
2^x = k = 5
2^x = 5
x = ln(5)/ln(2)
x = 2.321928
----- check -----
8^x + 2^x = 130
8^2.321928 + 2^2.321928 = 129.999997
----- final results -----
■ x = ln(5)/ln(2)
■ x = 2.321928
🙂
too long , when you let k = 2^x,
let f(k) k^3+k-130=0
just find the factor of f(k),
that's 5 ,(when k=5,f(k)=0),so that k-5 is the factor of f(k),
use the long division method,
we get
(k-5)(k^2+5k+26)=0
k=5,k= (-5+/- sqrt(79)i)/2
2^x =5
x=log(2.,5)
x=2.32 (3 con fig)
@@anson2075same thing really, cardano is more general to apply to other problem.
Though a faster is observe x=5 is a solution of x^3 + x = 130, because x^3 + x is strictly monotonic so x=5 is the only real solution
So no complex values??! Huh?
@@johnsmith1953x Hi Sir, Below is the continuation as per your expectation...
/// euclidean division
• from k^3 + k - 130 = 0 we got k = 5
• k^3 + 0·k^2 + k - 130 = 0
• (k^3 + 0·k^2 + k - 130) / (k - 5) = (k^2 + 5k + 26)
/// simple quadratic equation
• k^2 + 5k + 26 = 0
• Δ = 5² - 4·1·26 = 25 - 104 = -79
• √Δ = ±i√79
-> root #1: k = (-5 + i√79)/(2·1) = -5/2 + i√79/2
-> root #2: k = (-5 - i√79)/(2·1) = -5/2 - i√79/2
/// root #1: k = -5/2 + i√79/2
1) mod of k:
• mod = √[(5/2)^2 + (√79/2)^2]
• mod = √(25/4 + 79/4)
• mod = √(104/4)
• mod = √26
2) arg of k:
• arg(k) = π - arctan(√79/5)
3) exponential form of k:
• k = √26·e^i·(π - arctan(√79/5))
4) recall: k = 2^x:
• 2^x = k = √26·e^i·(π - arctan(√79/5))
• 2^x = √26·e^i·(π - arctan(√79/5))
• x = ln(√26·e^i·(π - arctan(√79/5)))/ln(2)
/// root #2: k = -5/2 - i√79/2
1) mod of k:
• mod = √[(5/2)^2 + (√79/2)^2]
• mod = √(25/4 + 79/4)
• mod = √(104/4)
• mod = √26
2) arg of k:
• arg(k) = π - arctan(√79/5)
3) exponential form of k:
• k = √26·e^i·(arctan(√79/5) - π)
4) recall: k = 2^x:
• 2^x = k = √26·e^i·(arctan(√79/5) - π)
• 2^x = √26·e^i·(arctan(√79/5) - π)
• x = ln(√26·e^i·(arctan(√79/5) - π))/ln(2)
/// final results
■ x = ln(5)/ln(2)
■ x = ln(√26·e^i·(π - arctan(√79/5)))/ln(2)
■ x = ln(√26·e^i·(arctan(√79/5) - π))/ln(2)
🙂
Oh yeah sure because everyone remembers the cubic formula just like that. 😂
I just brute forced this knowing that 8*8 is 64 and 8*8*8 is 512 it would have to be between 2 & 3 then via estimation and guess and check 2.32 came the closest in approximation. Even though I know this method is wrong on a standardized test with multiple questions and limited time. It is effective having a “feel” for numbers and proportions gets you through most of these tests
It is exactly how a physicist would calculate it
@@alonsobruni8131 **This is the way**
You're either into physics or computer science. Mathematicians look for the exact symbolic value.
Physics and chemistry…I’m algorithmically challenged 🤣
Used same method in about 2 minutes. Might have been able to do it analytical 40 years ago.. Hardly used any analytical math since poly technical. It's now a distant memory.
From the line y^3+y-130=0
(y^3-125)+(y-5)=0
(y-5)(y^2+5y+25)+(y-5)=0
(y-5)(y^2+5y+26)=0
This way it took me 3 extra lines to get to the same result that the admin here took 6 lines.
While watching this video at 2 AM, my brain slowly dies...
this is brain nourishment
SAME!!! 2AM TOO!!
Same here😂
1:37 😢
I am watching at 5.37 with red eyes 🫥
At the second step, if you can see that 2^x = 5, you have the answer, and you'll have time to answer some other questions. These questions usually rely on having an insight like this, rather than on grinding out a solution, so you should be asking at this point, "is there a cube slightly less than 130?"
A guess is a guess. Without demonstrating thinking process, replacing y = 26y - 25y is nothing but a guess, which teaches no general approach to solving similar problems. It is no better that noticing that y^3 + y grows pretty fast, so we only need to try a few integers to get 130. And 5 pops up almost instantly. It’s also a guess, but much more natural and less contrived.
agree with you, applying y = 25y - 25 y is far from a demonstration process, it is simply a trick. I prefer to use my brain and notice, as you did, that y^3+y grows fast
Isn't that what mid term split is all about tho? We deliberately choose those factors that gives the constant on multiplying and the coefficient of the lesser degree variable in the equation on adding?
equations can have irrational, complex and non integer roots, which are often impossible to guess (although they can be approximated). What you made was a guess for an integer root, and its standard practice to check a cubic for an integer root first, and then use factor thereom. Sometimes, an integer root does not exist or is too large and you must use cleverer methods. You often need to find the complex roots (which were skipped over by the poster)
It is not contrived to guess a particular factorisation. A lot of good algebra problems are about making clever observations, factorizations, splitting etc. Many good problems asked in olympiade and competetive exams all around the world are based on using clever little observations, which instantly simplify the problem.
There are 'non contrived' ways to solve cubics, but they are absurdly long (check out the cubic formula. The derivation is pretty interesting). And they do not even exist for a polynomial equation greater than 4 degree.
Any integer root would divide 130. And because y > 0 we only need to check the positive ones.
Remember this being taught in school, when I was about 13-14 years old. Never once used it. Forgot it completely. Then, at age 43, I needed to pass an exam for a certain university and I was taught such things again. Never ONCE used it. Forgot it completely.
This is like learning a language, that only few people speak for real in the world. I was taught several languages in school as well. Didn't know it at the time, but that has proven to be excellent usable knowledge, that I literally use each and every day in my normal life.
The history lessons keep from falling for just any propaganda. The geography lessons gave me basic understanding of who's where in the world, science and biology brought some insight too. But THIS? Completely superfluous. An utter waste of time. This ought not to be taught in schools.
From 2^x=5 follows x=log5, base2. It is by a definition of logarithm!
that's what I thought!
@@orlevene9964 Not normally left like that. The day will come when you have to use a calculator without a base 2 log function, so it is best to know how to change to base 10 or natural log.
I have never felt so bad at math in my life as I did trying to figure this problem out. Wow!!
As soon as I saw that we had a cubic to solve, and 130 has a prime factorisation to 3 primes, 2x5x13, I knew one of them was going to be a candidate! Also if you dont do the 26-25 trick, and consider what the coeficients of general cubic would be, given there is no y^2 coeficient, you get the property that the sum of the roots is 0, and the product of the roots is 26.
My approach before Watching this video
put 2^x=t
Now, we have a simple cubic equation
t³+t=130
=>t(t²+1)=130
Now,by trail and error method
Factors of 130 = 2×13×5
Putting t=5 in LHS,we get 5(25+1)=5×26=130
Hence the value of t=5
As,2^x=t (Taking logarithm both sides)
=>xln2=lnt
Put the value of t=5
x=ln5/ln2
Got the answer by simple algebra manipulation and took 2 minutes to write and seconds to solve
It is much easier to follow also :)
Once you have the cubic in t, just try looking for a perfect cube that's not much less than 130.
Trying 125 = 5³, we have the solution, because 5³ + 5 = 130. The rest follows as you describe.
Fred
Yup you just need basic 10th grade mathematics fundamental concepts and you can solve it easily within 3 minutes This made me realise how hard Jee Advance is.
I'm impressed to see so many alternative solutions. I see that the followers of this channel are true mathematicians. Well done! I would have solved the cubic equation using synthetic division. Best regards from Bogota.
A middle schooler with a calculator: Nah Id win
I don't think you can do this with a calculator? whats the approach here
@@vikrantsingh6580you can actually with a sceintific calculator
not funny and you cant do it with calculator either
@@raheelkhan-ek9qe you actually can bruh
@@stannate how then bruh
I'm starting to freak out that after reading most of the comments I can't find any criticism of the line at 3:43 that claims
y^3 + 25y = y (y^2 - 5)
What am I missing here?
I had paused the video to look for comments. After continuing the video all is well, and I am no longer freaking out.
@@jonnamechange6854 Ha ha, so much anguish, happened to me too... and it took just a couple of seconds of patience to find salvation!
Yes - that gave me a WTF moment until he corrected himself a line or two later.
By inspection, 2 to the x = 5. Hence x = ln5/ln2
2:25 i didint get it why you write y=26x25
yes also didn’t understand whats tge relation between 130=26*5 and 26y-25y?
At 3:30 factoring out y from y^3 -25y how does this become y(y^2-5) not y(y^2-25) ?
corrected 10 seconds later....ok.
Using bracketing (artillery method for finding range of target) and trial and error method should be able to do it in one or two minutes which would be much faster and more importantly less prone to algebraic error, but of course less elegant. Also if it is a multiple choice test you could quickly try the different answers and find the correct one in very little time.
Muy interesante video, muchas gracias por compartir tan buena explicación. 😊❤😊.
You could substitute y=2**x then use the rational root theorem to find the factor (x-5)
Much better.
thats what I did lol
Let us replace 2^x by X, then X^3 + X = 130, and evidently X = 5 is a solution.
As the function X -> X^3 + X is strictly increasing on R+, the value 130 is reached by the single X = 5.
So the unique real solution is given by 2^x = 5, i.e. x = Log 5 / Log 2. Thank you for your videos !
I suspect that Cambridge tripos candidates would be expected to consider the complex solutions as well.
8^x + 2^x = 130. 8 = 2^3 so we have 2^3x + 2^x = 130. You can say, immediately, that by inspection 2^x = 5 because 5^3 + 5 = 125 + 5 =130. Thus x = log5/log2 . I can't do that in my head - I know it's 2 point something and my best guess would be around 2.3 - but a few seconds on a calculator tells me that x = 2.3219 to 5 sig fig. Or if you need a little more time you can try 2^x = y so y^3 + y = 130 and again you can immediately see that y = 5. The rest follows - no need for all that manipulation for a simple problem which is what, I think, the Cambridge examiners would want to see. No need to use a sledgehammer to crack a peanut.
Interesting video, but if you're going to write the product as 26*5, why not just write it as 25*5 + 5 since you have y^3 + y, so 125+5 = y^3 + y, 5^3 + 5 = y^3 + y therefore y = 5 x = log_2(5).
By observation from y^3+y-130=0 we get y=5 so x=(ln5/ln2) which verifies the initial equation
8^x + 2^x = 130
(2^3)^x + 2^x = 130
(2^x)^3 + 2^x = 130
u = 2^x
u^3 + u = 130
u^3 + u - 130 = 0
u=5 is a root (u^3 = 125, 125 + 5 - 130 = 0). So,
5 = 2^x
log(5) = log(2^x)
log(5) = x*log(2)
x = log(5)/log(2)
This does indeed solve the equation. There would be another pair of solutions, which we could get by factoring x-5 out of the cubic and solving by usual quadratic methods.
Lets write the complex roots, as well.
y²+5y+26=0 ⇒ y=√26·[(-5±i√79)/(2√26)].
To evaluate log of y, we choose the branch cut of the log to be the non-positive real half-line & restrict the arguments of the complex numbers to lie between -π & π. Then,
log₂(y)=log₂(√26·[(-5±i√79)/(2√26)])=log₂(√26·e^(±iθ)) for θ=cos⁻¹(-5/(2√26))≅.663π
⇒ log₂(y)=log₂(26)/2±iθ/ln(2) where "ln" is the natural log.
2:00 Just leave the 130 on the RHS and factorise the LHS:
y (y^2 + 1) = 130,
130 = 5 × 26 = 5 × (5^2 + 1)
y = 5.
What an absurd long solution. Thanks I never thought of going to cambridge but earning a lot of money without killing my brain. :-)
What do you do
@@tegathemenace
Flexing, that's his job and it doesn't pay well
ln5/ln2 is the ans. I am currently in 11th and i did this by basic algebra.
Assume 2^x = t and then youll get a cubic eqⁿ. One of its roots by inspection is 5, and the other remaining Quad. Eqⁿ is a always +ve factor, so it doesnt matter. Hence there is only 1 solⁿ, y=5
=> 2^x=5, ln(2^x)= ln5, x(ln2)=ln5
x=> ln5/ln2 or log5 base 2👍🏻
You don't need natural logs. Log base 10 works perfectly as does any base. ln isn't wrong but you don't need it.
Let y= 2^x
y^3 + y = 130
y = 5
x = log to the base 2 of 5.
There's a fair bit of luck in this approach but it genuinely took about 30s.
Algorithm gave me this after looking up beginner algebra for my 7 year old (she wanted to know!).
I haven’t done this math in, oh, 10 years and some of it before then. So while everyone else can spot the obvious based on the comments, I really did enjoy the extra explanations. I could follow fairly well despite not having used those neural pathways in a very long time.
Perhaps if someone had done so when I first leaned this math, I’d be an MD instead of an attorney!
Oh okay I see you went for the longer method. The moment I saw the question the intuition was very clear because of exponential I immediately used log on both sides, and put the values to the calculator.
Using log on the left hand side does not help you in any way for solving this equation.
Newton’s method gives an approximation of log5/log2 by nine significant digits. Super cool problem!
The answer to the title card is: (8 x 13)+(2 x 13)=130
X=13
Technically it is simple division but you get there by multiplication. You take the 130 as your cap and see how many times you can multiply the 8 to fit in there. While leaving room for the 2 times whatever, by repeating the process for the 2 and adjusting the 8's multiple as needed you'll get to the point the that you're multiplying both numbers by the same ammount to fill out the cap. Which is what you want in this case.
8x + 2x = 130
( 8 × *X* ) + ( 2 × *X* ) = 130
( 8 ×13 ) + ( 2 × 13 ) = 130
8 × 13 = 104
2 x 13 = 26
104 + 26 = 130
X = 13
Just eliminate the common factors x and do the math... SIMPLIFY...
At 2:12 you missed a huge opportunity to use the Rational Root Theorem. If you had used it, you could have shown right away that y=5 was a root. You could then have used polynomial division to find the other factor as y^2 + 5y + 26. Instead you applied 26y - 25y as magic without any justification. This does not help the student.
If you like base-10 logs, log(5) = log(10/2) = log(10) - log(2) = 1-log(2), so you get x = 1/log(2) - 1. We know log(2) = 0.30103, so we can quickly compute it.
7:40 Since you've mentioned complex solutions, 2^x=5 has a foliating solution in ℂ, just like any logarithm. By the fundamental theorem of algebra, there are 3 solutions in ℂ to the cubic polynomial in y, so there were not even a need to compute D to find out that there are 2 more solutions for y in ℂ\ℝ. That there is only one y∈ℝ⁺ is clear if you rewrite y³=130−y and consider both sides as functions ℝ⁺→ℝ⁺: [y³=f(y)]=[g(y)=130−y]. Both f and g are continuous and surjective, so f=g has at most one solution. Note that f(y) is also monotonically increasing, g(y) decreasing in y, so there exists a solution. The other two got to be complex.
Just for the fun of it, since 2^x=exp(x log 2), the solution to exp(x log 2)=5 is foliating in ℂ: x=(log 5/log 2) + 2πki/log 2, k∈ℤ. x∈ℝ only for k=0, obviously. The complex roots of y²+5y+25=0 are not only extremely ugly, as you correctly noted, but also ought to foliate with the same period 2πki/log 2.
Miss those jee advanced days. Glad to see our students are much advanced, if this is the standard of the paper
How come no Nobel Prize winners or Fields Medallists has ever come out of any IIT ?
The first thought is to replace 8 with 2^3, then cubic equation follows. We do not expect kids to remember Tartaglias formula so this is a dead end. If we hope for a integer solution it is good to remember that the free term (130) is a multiplicative of all 3 roots. Indeed it can be factired as 2*5*13. By checking al divisors of 130 (2,5,10, 13, 26, 65) it turns out that 5 is a solution and it is easy to see that it is also the only real solution.
I really anticipated the iterative methods of numerical methods,Jacobi iteration and Gauss-sidel
This is scary, I seriously respect people who understand math . What a torture ?!
It's basic algebra that every single middle schooler learns
Not necessarily. I never had this in highschool
If you use prime numbers, the equation is easy to solve. That is 2^3x +2^x=(2)(5)(13) but this can also be written (2^x)(2^2x + 1)=(5)(26). And therefore 2^x=5 and x=log2(5).
Yeah all that was my first thought too. Easy work 😶
Being a middle school student i thought that this is an algebra equation in one variable😂😂. I thought it is 8x + 2x = 130 and here x = 13 i was like it is so damn easy why are they freaking out😂😂😂
Don’t feel bad I thought the same, but I’m a middle aged idiot! You still have a chance 😊
u³+u = 130. Think of a number whose cube + the number = 130. Obviously 125+5=130, so 2^x = 5. Hence x = log5/log2 = log₂5. Very easy. Did it in my head from the thumbnail!
Just substitute 2^x as t and solve it. Done. Wow! If this is one of three harder questions of the paper, i can crack it.
Nice problem. It would be more helpful if you explained why the substitution y = 26y - 25y would actually occur to you at the step you take it, rather than just show that "it works out nicely". Why not instead notice that 130 = 13 x 10, and proceed with y = 13y - 12y? I.e., explain that because 25 is a perfect square you'll get a factoring present in a moment.
Just do prime factorials of 130 = 5^2 * 2 * 3
Once you have y^3 + y = 130 =
y^2 * (y+1) you now need to remember that 2 * 3 = 5 + 1 and you’re done
I gave this problem to the new ChatGPT model (o1) and it got the same answer and solved it in a very similar way (although without the 26y - 25y stuff). Impressed!
It's really easier to solve when you already know the answer.
There should be 3 complex roots if the principal value of ln z is used. One of which is the solution presented.
I tighten bolts and signal cranes for a living and got this inside of about 3 seconds. Cambridge should really tighten up their entrance requirements.
Well your overqualified for your job then and should look for a job that requires more intellect than tightening bolts because only 1% of adults in the general population could solve this problem.
You tighten bolts and signal cranes for a living, but that doesn't mean that you aren't a math genius too!
@@charlesstimler9276 Arithmetic, geometry, and algebra maybe. I have zero clue about calculus
y=5 because life's short and when faced with y^3+y=130 you check what you can before being forced to argue as you must. Are there other relevant solutions to the cubic? Picture the graphs of 2^x and 8^x for real x. On the LHS of the origin we have a small sum,
Easy ques and i m going to Cambridge...wish me luck❤
I have graduated in Mechanical Engineering from RUT in Poland. 32 years ago. And I solved this problem in three lines. :) Applying logarithms.
Good Luck! :D
@@piotrstrzelczyk5013 Thanks mate :)
@@piotrstrzelczyk5013 same but i keep getting 2.
I changed 130 to 2⁷ + 2. Changed the other side to powers of 2 as well
Took log, then cancelled
And ended up with 4x = 8
When I got my first programmable calculator, an HP25, I wrote a simple bisection routine to solve the countless equations I came up with when taking physics exams. That is still how I would choose to solve problems like this. I leave math to the mathematicians.
Professors, who had been raised at a time when 4k of memory had to be rented because it was too expensive to buy, would tell me my approach was too expensive. I think about those comments every time I look at my stash of gigahertz laptops that are collecting dust.
Once you have a cubic, you have a known algebraic solution. Finding clever ways to extract the root quickly after this point is just a Sudoku puzzle.
It's comes to 8+2 = 10. X is 13. I failed algebra and did this in my head. 104 + 26.
The original question hides the real question: the magnitude of difference between two cubes.
8^x + 4^x + 2^x = 155. Same idea. Same answer.
Interesting.
The solution hinges on writing 130 = 5 x 26, which is noted and exploited in the on-screen "working out".
Similarly observed: 26 = 25 + 1 = 5^2 + 1.
Hence 130 = 5 x (5^2 + 1) = 5^3 + 5.
Comparing 5^3 + 5 = 130 with 8^x + 2^x = 130 suggests 2^x = 5 and 8^x = 5^3, the latter of which can be rewritten 8^x = (2^3)^x = (2^x)^3 = 5^3, confirming that 2^x = 5.
Finally, 2^x = 5 => x = log5/log2, as stated on-screen.
Log base 2 of 5 falls out by simple inspection! Why so difficult?
thankyou interesting method
I used to be amazed by calculus, I was never fluent, I was really bad memorizing all these little rules.
This feels like someone playing guitar really well in front of me but I'm into electronic... Like dude I wish I had the same passion for this subject.
Before watching (so I might be very ashamed after!)... but it seems too easy. The BIG clue is that 8 is a power of 2 (2 cubed). So 8^x is (2^3)^x. And then we need a bit of substitution, so y for 2^x. Which means we have y^3 + y = 130 and it's clear there are very few options (as cubes get big quickly, so 6 is already a bust). It (y) must be 5. so 2^x = 5 and 8^x = 125. x is a bit more than 2 (2.2 or 2.3 ish). so i need a calculator to get exact number. or just express as logs as 2^x = 5 is same as x log 2 = log 5 therefore x = log 5/log 2. which is 2.322. And, of course, 8^2.322 + 2^2.322 is roughly 130. (130.019)
Good work. But too long. When we arrive at the point 2^x=5 (at 7:37 moment), we have by definition
2^x=5 x= log 5 in base 2. Why so many steps?
Why do suppose Y^3 - 25Y equal Y(Y^2 - 5) instead of Y(Y^2 - 25)
This was so damn easy like I had started preparing more jee about 4 months ago, and still solved it, like this exam zeems so easy and nothing compared to jee if these type of questions come in it
And yet no Nobel Prize winner or Fields Medallists has ever come out of an IIT
I make this: 8^x+2^x=130
(2^x)^3+2^x=5^3+5
If we named 2^x as Y. Therefore, we have: Y^3+y=5^3+5
Which immediately means that Y=5
So, 2^x=5.
The equation show us that the answer is log 5 to the base of 2=x 😁
it is far long, its shorter answer
x^2(1-x)=12
and
12= 4*3
so 1-x=3
x=-2
and x^2=4
x=-2
so, the answer is (-2).
I’ll be honest with you I might be screwed I’ve done so many questions like these my first thought was “I’ll take the natural log of both sides” and my second thought was “maybe we can write in u as 3x my brain is so calculusified I can’t solve basic algebra
Eureka, my insomnia has been cured. (Fell asleep at 4 min.)
After substituting y, we could factor out y and find the product of 130.
y(y^2+1)=130
y(y^2+1)=26×5
y=5
y^2+1=26
y^2=26-1
y^2=25
y=+5,-5 but y=5 is the same as th other one.
Therefore since both y are equal then y=5
Then we can solve for x
But you cant just turn 130 into any 2 numbers you want to then set y equal to them multiplied together because there are infinitely many combinations. You could make it 2*65 if you wanted and get something completely different.
It is about choosing a the product that captures it the best
математикам известна одна единственная формула пифагора, и как только они ее не выкручивают чтобы получить целые области для одурачивания публики, при этом используют метод обратного реинженеринга чтобы прийти к заранее известному результату
8 EXP X = 4*2 exp x Re-writing the equa. 4*2 exp x + 2 exp x = 5* 2 exp x = 130; 2 exp x = 130/5; x log 2 = 26; x = log 26/log 2. Solve from there. Basically, 2 exp 6 = 128, so we're looking for x
(4*2)^x is not the same as 4*(2^x) though,
X = Log(130)/ Log8 + Log2 = 2.3
logs dont work like that brother
No
I think your approach is more complicated and time-consuming (since it's an entrance exam and the time is limited, we should consider the easiest and the fastest approach to minimize the probability of mistakes and spend the least amount of your time). So, for me, it was faster and simpler to make in this way:
8^x + 2^x = 130
2^3x + 2^x =130
2^x = p; x=log p [base2]
2^3log p [base 2] + 2^log p [base2] = 130
p³ + p = 130
p(p² + 1) =130
Now we can just pick a number, the calculations are not complicated. We need to find a number that, when squared, is not too big (because it will then multiply by itself). We can consider the cubes of numbers from 3 to 6 because they are quite easy to calculate in your head. 3³=27, 4³= 64, 5³= 125, 6³=216. The cube of 5 is closest to 130, so we can easily and quickly check that 5 is the answer for p(p²+1)=130 (I think it is not difficult to calculate 26*5 in your head). If you have had experience with quick mental arithmetic at school or with your tutor, then to pick up number in this equation will take you no more than a minute and a half.
p=5
x=log5 [base2].
Your thoughts
Where in the world, would you use this ability/facility in a practical sense?
The 26-25 was kinda crazy 😅. Who’d thought of that???
It is immedistely obvious that there exists exactly one real solution (the LHS is a monotonically increasing bijective function from the reals to the positive reals). Since 8=2^3 and 130=5^3+5 we have x=ld(5), where ld denotes the logarithmus dualis.
This is too simple to be an olympiad problem. Just take 2^x to be a variable 'a'. Now solve the quadratic and find the value of x
at 3:27, why do you factorize y^3-25y into y(y^2- 5), it should be y(y^2-25), did I get something wrong?
sorry, I continued, and he corrected,
When you have 8^x + 2^x you have something cubed + something. That makes 130 = 125 + 5, so the something is clearly 5. Then 2^x = 5 and x = log5 / log2.
Exponentials like these don't have multiple real solutions, because they are positive and monotonically increasing, so that's the only real solution.
That takes less than 30 seconds, not 10 minutes, and you really need not to be hanging about on a Cambridge entrance exam.
Love the y= 26y -25y out of nowhere step.
Who says that you are interested in only real number roots? It does NOT state that in the instructions (which actually do not exist anyway), and you never stated it at the beginning of the video.
Super 👍 from Algeria
As an indian i say . Math Olympiad questions are easy asf
hp 49g+ couldn't solve it symbolically.
It says "SOLVE Error: Not reducible to a rational expression"
However, the numeric solver did just fine.
Was going to use Cardan's method for cubic equations to solve that. This 26-25 is an interesting approach but may not be applicable for all sums😊
Xlog8 + xlog2 = log130
X (log 8 + log 2) = log 130
X =. Log 130 / (log 8 + log 2)
Problem solved?
9:14 Aug 4
One hundred and surty was the best part of the solution
Cambridge Entrance Exam: 8ˣ + 2ˣ = 130; x =?
130 > 8ˣ > 2ˣ > 0; x ϵ R, no complex or imaginary value root
8ˣ + 2ˣ = (2ˣ)³ + 2ˣ = 130 = 5(26) = 5(5² + 1) = 5³ + 5
(2ˣ)³ + 2ˣ = 5³ + 5, [(2ˣ)³ - 5³] + (2ˣ - 5) = (2ˣ - 5)[(2ˣ)² + 5(2ˣ) + 25] + (2ˣ - 5) = 0
(2ˣ - 5)[(2ˣ)² + 5(2ˣ) + 25 + 1] = (2ˣ - 5)[(2ˣ)² + 5(2ˣ) + 26] = 0; (2ˣ)² + 5(2ˣ) + 26 > 0
2ˣ - 5 = 0, 2ˣ = 5; x = log₂5 = 2.322
The calculation was achieved on a smartphone with a standard calculator app
Answer check:
x = log₂5, 2ˣ = 5: 8ˣ + 2ˣ = 130; Confirmed as shown
Final answer:
x = log₂5 = 2.322
Let's put a box around it. How exciting.
Why doesn't using base 10 logs work here? 🤔
Such as:
x( log 8) + x(log 2) = log 130
Then solve for x 🤯
It does not matter what base you use: log5/log2 will always give the same number.
I’m just doing basic algebra with my son for his year 7 homework. Why isn’t it simplified to 10x = 130? So x = 13? That’s what my son has just learnt to do.
Because it is 2 to the power of X, not 2 times X. Completely different equation.
After learning that, well i did 35 year ago, and much more, I am standing in the line of unemployed people for free food. How does it help me? Anyone?
sigh, as a mathematican, there are 3 solutions, log(5)/log(2) then log(-5/2+I*sqrt(79)/2)/log(2) and log(-5/2-I*sqrt(79)/2)/log(2) not one.
First thought: "Oh, that's simple, just 128 + 2, which is 2⁷ + 2¹"
Then after diving in: "Something between 2 and 3, if they chose =68 or =520 this would be simple"
Is there some hidden agenda in writing x like an egypt hyroghlyp inthead of just to lines crossing ? Ah I got it it takes longer so you have more time to speak during writing. ;-)
2::19 he says: "130 can be read as the product of 26 times 5" --- 130 = 26 x 5
--- WOW! UNBELIEVABLE! --- For a) 130 = 65 x 2 is the same as b) 130 = 13 x 10
*WHY* did he not choose a) y = 65y - 64y or b) y = 13y - 12y ???