I wasn't good at math. That said, solving this equation was like driving from LA to San Francisco and ending up in Toronto, Canada.
@robertmcdonald32977 ай бұрын
6
@robertmcdonald32977 ай бұрын
The answer is 3
@marol5594Ай бұрын
Nice solution, but way too long! Don't make your lifes harder than necessary: At time 02:30, just compare the coefficients: y^3 + y = 3^3 + 3 y = 3 2^x = 3 x = log3 / log2 (+ saved 10 minutes 😊)
@jropar135 ай бұрын
I don't think that anyone who was worried about the music has the patience for the math. Just my opinion. Thank you so much. That was a lot of work. When you're buried in the math, anything else should fade to oblivion.
@jim237610 ай бұрын
2^x = u. u^3 + u - 30 = 0. Factor: (u - 3)(u^2 + 3u + 10). Real solution u = 3. Quadratic on the second term for the two complex solutions. Back substitute. Messy problem.
@aimulus8 ай бұрын
If complex numbers are allowed as solutions, then there are many more solutions, namely x=(log(3)+2*k*π*i)/log(2) and x=(log(10)/2+(i*(±atan(sqrt(31)/3)+(2*k-1)*π)))/log(2) for every integer k.
@marianneoelund29408 ай бұрын
Complex numbers are always allowed in my world. However, this isn't a very interesting expansion of the solution set, as it just says that running around a circle an integer number of times will return you to your starting point. Exp[(2×pi×k + c)×i] = Exp(c×i) for any integer k.
@oahuhawaii21416 ай бұрын
Close, but no cigar. You mixed up the bases for the log functions, if the base for the log function isn't explicitly e. All the real components have a form of log(r)/log(2), regardless of base {such as 2, e, or 10}, as they all yield log₂(r). However, the imaginary component is log₂(e^(i*θ)), which is simplified as ln(e^(i*θ))/ln(2) = i*θ/ln(2) . You wrote the equivalent of i*θ/log(2), but this log function is traditionally base 10.
@aimulus6 ай бұрын
@@oahuhawaii2141Yes. The natural logarithm has to be taken, of course.
@jhpark6958Ай бұрын
참 힘들게 푸시네
@Daimler527 ай бұрын
Спасибо за познавательное видео уровнения такого уровня были когда поступал в технический университет в России в 1999 году. Тренировался считал десятками такие. Сейчас уже конечно подзабылись формулы, но с учебником думаю решил бы и сейчас.
@francescob.14123 ай бұрын
The thumbnail got me very intrigued, and I accepted the challenge. I solved it on my own, and I managed to find x = 3 in a much simpler way. I still applied substitution, t = 2^(x) , so x = log_2(t). But when I reached t^(3) + t = 30, i scomposed left and right side into t (t^(2) + 1) = 2*3*5 And 2*3*5 = 2 (3*5) = 3 (2*5) = 5 (2*3) , so I had to choose the only combination that would match the left side, and the only one that matches is 3 (2*5). In fact, 3 * (3^(2) + 1) = 30. So t = 3, and x = log_2(3) . But then, when I confronted it to the solution you gave, I realized that I found only the real solution. Was it sheer luck because logarithm is defined only for real numbers, or my solution had some sense?
@miguelorobitg20405 ай бұрын
One in top righ of eight, another one in top right of number two. I found two of them! Genius!
@StephenC0509 ай бұрын
That pen writes quite beautifully
@DreamyCheshire-up9rf9 ай бұрын
17/1/24 ( Wednesday ) 9.15 pm. Sorry, I failed maths in school. I thought it was a primary school algebra. Never thought that in order to find the value of X, you need to create a value of Y with log included, which I could not comprehend. I was wrong to think that the value of X was 3. Instead, after so many steps, the value of X was another equation.....no wonder I failed Maths.
@dak718898 ай бұрын
X value will be between 1 and 2. Simple solution. Lets take x value 1 then you will get 10 , let’s take x value as 2 now you will get 64 plus 4 that is 68 . 30 is between 10 and 64 so x value will be between 1 and 2. In India we get such questions as multiple choice answer questions and we get less time to solve.
@alexxander66567 ай бұрын
This was definitely the long way in my opinion. I definitely forgot the other way tho. Too many things you just have to know like 2^3 is 8 and breaking 30 in to 27 and 3. M.
@MrHotlipsholohan7 ай бұрын
You are a person with a good knowledge of the laws of indices , and logs, loved all this at secondary school 45 years ago bit have forgotton some of the rules so get lost near the end , lol, it pays to know these rules and got me out of many a problem in my engineering career , esp algebraic rules
@estifanosberhe23214 ай бұрын
You are a genius😊by the why I'm estifanos' son. I'm going n grade 5 and I enjoy your math videos.😊
@troubledsole91047 ай бұрын
Knowing the theorems was essential to solving for X.
@paulssamuel9 ай бұрын
let y=2^x then y^3 + y = 3^3 + 3 => y=3 => x = log3/log2
@praiseYAHalways7 ай бұрын
soooo X = 3 as in 8x3 (24) plus 2x3(6) equals 30???? ....... at least that's how I did it in my head....took about 5 seconds maybe less 🤔 unless I'm reading the problem wrong...
@kwachitshanbhag1974Ай бұрын
@@praiseYAHalways wrong. 8 cube and 8x 3 is different
@iangray54077 ай бұрын
I'm amazed at how many people get an answer of 3 while completely misunderstanding the nature of logs or that x is a power. The right answer for the wrong reason is the wrong answer.
@MounikaMix7 ай бұрын
Can we not directly apply log to the main equation and solve
@debkantadas76827 ай бұрын
But how can you get a 14 minutes video?
@iangray54077 ай бұрын
Simple answer: no. You'd still have to solve log(8^x+2^x) = log(30)
@pmacgowan8 ай бұрын
Well now I know why I just did applied maths, not pure !!, well done
@BarrieHughes9 ай бұрын
You could apply the log formula right at the start.
@marianneoelund29408 ай бұрын
log(x + y) = ?
@ezrapotter46319 ай бұрын
Right off the bat, x
@oahuhawaii21416 ай бұрын
What about a lower bound? Can x be negative or zero? Perhaps you can plot y = 8^x + 2^x - 30 to find the zero crossing(s).
@jcolivie9 ай бұрын
Is this the easiest way? So I prefer to study integral calculus, differential calculus, theoretical physics because they are much easier than that (contains irony).
@MathHero2412 күн бұрын
let u = 2^x. Then u^3+u =30. derivative of u^3+u is 3u^2+1 which is positive. Therefore, LHS is increasing and the equation has at most one solution. u=3 is an obvious guess. x = log(3)/log(2) is the only solution then!
@luckyuliveHI7 ай бұрын
Wow, I'm in finance and have used algebra in many situations. I don't think I'd have any use for this one. I wasn't expecting the solution to be this long.
@oahuhawaii21416 ай бұрын
Howzit! Monthly mortgage computations are messy, and often require high-precision, iterative calculations and adjustments due rounding to the nearest cent or penny, and lots of decimal places for simple or compound interest. The calculations are messy because the idea value doesn't fit to a perfect dollars and cents value. This introduces errors that must be dealt with in an empirical manner. I wrote a program to find the monthly mortgage payment. It finds the idea value, then rounds that and recalculates every monthly balance so that the final payment for each year takes into account the accrued errors. The math we use in these videos don't have these empirical aspects, and are "cleaner" to calculate.
@luckyuliveHI6 ай бұрын
@@oahuhawaii2141 hmmm. sorry brah, have to disagree with "high-precision, iterative calculations". Any online loan calculator will get you close enough, +/- $50 or so. Excel's native loan ss will also do the job... entered the basic loan figures from mortgage brokers and I'm always within $10 of their proposed payment.
@sergeyshamshura656110 ай бұрын
Very annoying music, please change it. I wanted to quit watching right in the middle of the solution
@joelo55510 ай бұрын
Mute it .-.
@TheSimCaptain10 ай бұрын
@@joelo555 If he couldn't figure that out, he probably can't follow the math. 🤔
@FuriousMaximum10 ай бұрын
I advise you to turn the sound off. Pretty simple.
@antonioherron38169 ай бұрын
Right please remake silent 🔇
@edmondscott74449 ай бұрын
Agree. Music is dreadful.
@57Jimflynn3 ай бұрын
Why do y sub instead of take the natural log or log of each piece immediately? xlog8 + xlog2 = log30 and go from there? X(log8+log2)=log30 X=(log30)/(log16)
@57Jimflynn3 ай бұрын
Symbolab gives your result. Butis it a matter of separating exponents?
@57Jimflynn3 ай бұрын
Is there an order of operations I'm not seeing?
@flyingmonkey57877 ай бұрын
This seems like a very complicated way instead of using log. How many identities did the op use? This feels like an algebra 2 video
@siddharth.2468 ай бұрын
Dude you could've just applied the remainder theorem when you had the equation y^3 + y = 30 or y^3 + y - 30 could only happen when y=3 and then applied the logs.
@stephenremo920010 ай бұрын
Like you can solve in 5 seconds in your head ... Or 10 pages of math
@Truthnvision6 ай бұрын
Wonderfully explained
@MathEducation100M7 ай бұрын
Nice trick
@ОльгаПчела-р3б8 ай бұрын
Тот случай, когда ответ, состоящий из логарифмов, оказался сложнее, чем изначальный пример.
@IrinaI55 ай бұрын
Как я любила их решать в школе!
@irina44478 ай бұрын
(8 +2)*=30 10 *=30 30:10=3
@KeithDCanada7 ай бұрын
I'm sorry, but the answer is obviously "pizza"
@formause12 ай бұрын
(8X+2X)=10x=30...-> /÷10..-> x=3
@phishmastermike7 ай бұрын
Simple 8 plus 2 is 10 10 times 3 equals 30 so x equals 3 To check it 8 times 3 is 24 2 times 3 is 6 24 plus 6 is 30
@kintagrama5 ай бұрын
Also a²+b²≠(a+b)²
@marcosindici59948 ай бұрын
Better than ASMR
@mdmu_vjogdand7 ай бұрын
So, what is log 2(3) ???? Can't we have a straight answer in numbers??? 😋
@ianclarke49457 ай бұрын
Y are you pointing at the rabbit? Because we are now going to follow it all the way down!
@usiseme4 ай бұрын
Gets harder as we go
@wizard105able10 ай бұрын
At 4.28 where is 1 come from? I lost it there.
@wizard105able10 ай бұрын
Ok nevermind after watching another video I understand it now
@cornerboytep9 ай бұрын
Where? I am confused
@tomref40019 ай бұрын
I lost it there too-whats the explanation for the 1, or would be grateful for the other video URL please?@@wizard105able
@kintagrama5 ай бұрын
@@cornerboytepWhen he converts y-3 into a 1
@Ninjaknight818 ай бұрын
Find X? There's two of them on the right side of the eight and the two😂 always remember it's usually better to be a smartass than a dumbass😂
@oahuhawaii21416 ай бұрын
After grading, you'll find another one covering your answer sheet.
@perry39288 ай бұрын
How about 8 Sq by 2, 2 Sq by 7. 16 +14 =30
@Amps_pic8 ай бұрын
เก่งมาก😊
@carlw8 ай бұрын
I get it except for how (y-3) became 1 at 4:27
@marianneoelund29408 ай бұрын
(y - 3) is a common multiplier and is being factored out to the left. Reverse distributive law.
@carlw7 ай бұрын
@@marianneoelund2940@marianneoelund2940 Thanks, I understand that, but no other (y-3) on the left is factored/cancelled. I get the 9+ 1 afterwards but.......
@marianneoelund29407 ай бұрын
@@carlw It's straightforward distributive law: (y-3)*z + (y-3) = (y-3)*(z+1) Just factor out (y-3). Nothing is being canceled.
@carlw7 ай бұрын
@@marianneoelund2940 Yes, it's been 40 years since doing that, forgot that. And to think I was doing integral calculus now forgetting stupid things like that. Thanks.
@Gizmo97077 ай бұрын
That's the real solution, but there are also complex solutions
@oahuhawaii21416 ай бұрын
Yes, an infinite number of them.
@hasanhumayun96007 ай бұрын
Make in math octal
@ahmedberi8 ай бұрын
Soruyu çözerken onluk log yerine ln de kullanılabilirdin daha şık estetik bir çözüm olurdu
@MacTaipan8 ай бұрын
Hm. I think after substituting y, splitting 30 into 27+3 will only occur to you if you already know the solution.
@VaNSaNT858 ай бұрын
Not really, actually that simple trick is a tool you use lot in calculus
@TaharOubagha5 ай бұрын
Esuatt start oddition.after result free Master combine
@Petirmenggelegar8 ай бұрын
👍👍👍
@ZsoltPal230920114 ай бұрын
thanks :)
@misterenter-iz7rz9 ай бұрын
2^x=3, x=log 3/log 2.
@danieloliveira50059 ай бұрын
Very good
@twinmomsurvival57448 ай бұрын
X equals 3, did it it 15 seconds. Love being a math geek :)
@marianneoelund29408 ай бұрын
I think you meant 2 raised to the power X is 3. But you missed the complex valued solutions.
@Senectus7 ай бұрын
I looked at the first page and intuitively made the same conclusions as @twinmomsurvival5744 can you explain how your answer is correct?
@marianneoelund29407 ай бұрын
@@Senectus The equation left side is 8 raised to the power of X plus 2 raised to the power of X. It is not 8 times X plus 2 times X. Read carefully. The X is superscript.
@iseeyou87817 ай бұрын
Nope.
@praiseYAHalways7 ай бұрын
@@marianneoelund2940 oh, are you saying that is 8 to the power of x plus 2 to the power of x....is that how it is written?
@TaharOubagha5 ай бұрын
Its a nice sweetch lieder number for churche master.29+(9)XYC=23(8)+2(/29)__3#CH>133IPS=122+(N14)+S128
@johnnyo59938 ай бұрын
The judges say yes the answer is 3 they roll big joints too
@mauriziograndi17509 ай бұрын
X=1.585 by inspection without writing anything, a couple of approximations were enough in about 3 minutes. Anyway congrats for all your effort.
@oahuhawaii21416 ай бұрын
x = log₂(3) = log(3)/log(2) ≈ 1.5849625007211561814537389439478 Note the "approximately equal to" symbol vs. the "equal to" symbol.
@jacekplacek82749 ай бұрын
y^3+y=30 3 is ok so (y-3)*Q=0 and then easy to the end
You would be correct if the question were 8x + 2x = 30 8(3) + 2(3) = 30 24 + 6 = 30 x = 3 That's not the question being asked here though, rather it's 8^x + 2^x = 30 Let's see what happens when we plug-in 3 8^(3) + 2^(3) = 30 512 + 8 = 30 X obviously does not equal 3 It's deceptively a much harder problem than it looks. The only thing I knew going into this was x > 1 8 + 2 = 10 x < 2. 64 + 4 = 68 After having watched this video, that's still pretty much all I know, so don't feel bad if this went over your head because I'm right there with ya.
@oahuhawaii21416 ай бұрын
@@wargrunt42: Here's the full answer, including all the complex solutions. 8^x + 2^x = 30 (2^x)^3 + 2^x - 30 = 0 Let y = 2^x: y^3 + y - 30 = 0 (y - 3)*(y² + 3*y + 10) = 0 y = 3, (-3 ± i*√31)/2 For y = 3: 2^x = 3 x = log₂(3) = log(3)/log(2) ≈ 1.5849625007211561814537389439478 Actually, e^(i*π*2*k) = 1, for integer k, so we can generalize: 2^x = 3*e^(i*π*2*k) ln(2^x) = ln(3*e^(i*π*2*k)) x*ln(2) = ln(3) + i*π*2*k x = (ln(3) + i*π*2*k)/ln(2) x = log₂(3) + i*π*2*k/ln(2), k integer For y = (-3 ± i*√31)/2: 2^x = (-3 ± i*√31)/2 = r*e^(i*θ) { polar form } r = √((-3)² + (√31)²)/2 = √10 θ = π*(2*k+1) ± atan(√31/3), k integer ln(2^x) = ln(r*e^(i*θ)) x*ln(2) = ln(r) + i*θ x = ln(r)/ln(2) + i*θ/ln(2) x = ln(√10)/ln(2) + i*[π*(2*k+1) ± atan(√31/3)]/ln(2) x = log₂(10)/2 + i*[π*(2*k+1) ± atan(√31/3)]/ln(2), k integer
@kintagrama5 ай бұрын
Dumb
@harkjohnny8 ай бұрын
I found 3 of them! Directions were in the preview image… one is black and two are red! 🥸
@ronaldnoll324710 ай бұрын
OK... I don't have a pure math degree. I'm not happy with the result, because X is a number and the result is still an equation. I'm really confused.
@FuriousMaximum10 ай бұрын
Thats because (-3 +- sqrt(31)i) / 2 Cannot be solved, thus it can't be simplified.
@stephenremo92009 ай бұрын
Doing some quick math in my head the answer is just under 1.6.. So it probably doesn't have an answer really 1.55 isn't enough 1.6 is too much
@marianneoelund29408 ай бұрын
The result equation contains only constants, thus it is final, even if it appears messy.
@oahuhawaii21416 ай бұрын
The expression on the right hand side can be evaluated, but it's an irrational number: x = log₂(3) = log(3)/log(2) ≈ 1.5849625007211561814537389439478
@SANTANUMAHTO2 ай бұрын
Sir pen ka naam Bata do plz ❤❤
@janetedossantoscardoso84449 ай бұрын
Muito bom
@maurindenouden82459 ай бұрын
3x 8 = 24, 3x 2 = 6. 6+24 = 30
@cutecat9867 күн бұрын
X= 1.59 (checked with calculator)
@ВасилийПупкин-ж2с7 ай бұрын
Меня всегда интересовал вопрос : "а нахер это всё надо?: Километр бумаги истратили
@chestermarcol38317 ай бұрын
Why wouldnt it be: x= log10(30)? which is x=30
@rajinibommagani63147 ай бұрын
I have major doubt atleast 8 square chesina 64: 64+some number=30 How is it possible
@wrippley1037 ай бұрын
It's right there above 8 and 2.
@oahuhawaii21416 ай бұрын
They raise those bases to the power of x.
@JakubSK9 ай бұрын
Did it faster in my head without all the extra work.
@parandersson65419 ай бұрын
It took me 5 sec in the head
@Read-My-Post-Idiot8 ай бұрын
That's a claim you will never be able to prove,
@PapaPoohBear9628 ай бұрын
It's simple. Take the x's out, and you get 10.
@jeffrobuck63389 ай бұрын
Simplification??
@colinmccarthy79219 ай бұрын
There must have been at least 15 steps to follow to find the solution. If that is simplification,I am quite astounded. Surely there is another way.
@mauriziograndi17509 ай бұрын
Yes there is another way
@laurier34057 ай бұрын
I got a 99 on my algebra regions but couldn't add anything today. 😅
@PO6OKOHb7 ай бұрын
Я не математик но (8*3 = 24) + (2*3 = 6) = 30, я не понял огромной формулы но ответ верно высчитал
@hijirikawa-chan74497 ай бұрын
Это студенческий пример, ты не вытянешь. Ты не отличаешь степень от произведения. И "x" равен не "три", а вот этому 7:03 Логарифм называется.
@johnballock92808 ай бұрын
Did I see a word 'simplification' , yet the final answer is ' not possible ' ....... Oh me, oh my... How dumb I am
@oahuhawaii21416 ай бұрын
There are 3 principal solutions for x. One is real, which is shown, and two are complex conjugates, which were discarded. In truth, there are an infinite number of solutions, and all but one are complex numbers. 8^x + 2^x = 30 (2^x)^3 + 2^x - 30 = 0 Let y = 2^x: y^3 + y - 30 = 0 (y - 3)*(y² + 3*y + 10) = 0 y = 3, (-3 ± i*√31)/2 For y = 3: 2^x = 3 x = log₂(3) = log(3)/log(2) ≈ 1.5849625007211561814537389439478 --- Actually, e^(i*π*2*k) = 1, for integer k, so we can generalize: 2^x = 3*e^(i*π*2*k) ln(2^x) = ln(3*e^(i*π*2*k)) x*ln(2) = ln(3) + i*π*2*k x = (ln(3) + i*π*2*k)/ln(2) x = log₂(3) + i*π*2*k/ln(2), k integer For y = (-3 ± i*√31)/2: 2^x = (-3 ± i*√31)/2 = r*e^(i*θ) { polar form } r = √((-3)² + (√31)²)/2 = √10 θ = π*(2*k+1) ± atan(√31/3), k integer ln(2^x) = ln(r*e^(i*θ)) x*ln(2) = ln(r) + i*θ x = ln(r)/ln(2) + i*θ/ln(2) x = ln(√10)/ln(2) + i*[π*(2*k+1) ± atan(√31/3)]/ln(2) x = log₂(10)/2 + i*[π*(2*k+1) ± atan(√31/3)]/ln(2), k integer
@ScorpioHR7 ай бұрын
This is 3x longer as it should be. And I guess the goal was to make it over 10 minutes mark. At one point she wrote exactly the same line as the one before, not a single thing was changed... 4:00 mark So,yeah, it's just milking the algorithm at this point. Despicable!
@kintagrama5 ай бұрын
What? Is not the same
@KenraMan-iw4ti7 ай бұрын
I found two X’s. One next to the 8 and the other next to the 2..
@oahuhawaii21416 ай бұрын
After grading, you'll have another one that's easier to find -- it covers your whole answer sheet.
@MichaelLong-k1u3 ай бұрын
3
@lizwirtz59687 ай бұрын
Is that supposed to be a 'simplification'? How does 'y' come into an equation that has no 'y' to start with?
@robbie3307 ай бұрын
The y is only a symbol, just like x. It’s easier to work with one variable (y) than to work with 2^x.
@lizwirtz59687 ай бұрын
@@robbie330 what is 2 ^?
@Abram_Israel_42359 ай бұрын
5:40 А зачем так сложно? Если 2^x=3. То x=log2(3). Это следует из определения логарифма. P. S Я не пользовался логарифмами больше 20 лет. 😅
@tz88118 ай бұрын
Вручную, подобрал сразу. Чего вола катает...
@ToxicTanker-w6o8 ай бұрын
Тоже хотел написать. Волосы дыбом от таких "полезных" манипуляций.
@_-Yan-_8 ай бұрын
@@tz8811 сразу видно что 3=x про логорифмы уже не помню, с этим перегнул!)
@khurshidchaudhry22279 ай бұрын
It's very complicated maths can't get my head around. Doesn't follow any particular rule.
@jeremyfogt67988 ай бұрын
Well you didn't get the -2 for not showing all your work.
@nickmacdx7 ай бұрын
Off course its the long way. Thr algorithm pays by the hour
@kintagrama5 ай бұрын
What?
@hanisitsobarna48975 ай бұрын
tease with cross ... 🤞 😅
@neo33739 ай бұрын
Rien que la deuxième ligne...? C'est quoi tout ces a, m, etc....?
@gilvansouza90918 ай бұрын
@MENSA.lady27 ай бұрын
Why make life hard. Let x=2 then it is obvious that x must have a fractional value somewhere between 1 and 2.
@vincenthalas70559 ай бұрын
I have a solution for cake failing to rise.
@samirraydan17129 ай бұрын
please. do quadratic (x-7) =3, tks
@marianneoelund29408 ай бұрын
Where do you want the square? It's missing from your expression.
@DeZ0077 ай бұрын
Complicated version. Surely you don't expect to make math interesting with your extended method?
@Mr_Misunderstood249 ай бұрын
Why do you show the problem, right “Solution” underlined, and then rewrite the problem? Seems redundant and a waste of time.
@VanOutloud8 ай бұрын
It's very simple algebra. Why complicate it? What number could (x) be to factor times eight, than times two add them together to equal thirty. keep it simple. (8x3) + (2x3)= 30, (24)+(6)=30.
@wargrunt428 ай бұрын
The question wasn't 8x + 2x = 30; that's a much easier problem to solve. The question was 8^x + 2^x = 30; if you simply plug-in 3 here you get 512 + 8 = 30, which is obviously incorrect. You can't just factorize 30 and call it a day here, the problem is way more complicated than that.
@garemacquarrie67058 ай бұрын
But it's still 3
@wargrunt428 ай бұрын
@@garemacquarrie6705 No, it's not. Do you not understand what exponents are? The question being asked here is 8 raised to the power of what plus 2 to the power of what equals 30? 8 raised power of 3 equals 512, which is way too much. X does not equal 3.