8^x+2^x=(2^x)^3+2^x=Y^3+Y=Y(Y^2 +1), with Y=2^x. On the other hand, 130=2*5*13=5*26=5*(5^2 +1). Therefore, Y(Y^2 +1)=5(5^2 +1)Y=5. Therefore 2^x=5 => x=log2(5).
Si ; 130 = 13×5×2 Entonces (8^X + 2^X) es multuplo de 13; 5 ó 2 Haciendo : 2^X = a 8^X + 2^X = 130 restando 2a a^3 + a - 2a = 130 - 2a a^3 - a = 130 - 2a a(a^2 - 1) = 130 - 2a a (a-1)(a+1) = (a-1)(a)(a+1) = 130 -2a y como todos sabemos... el producto de tres numeros consecutivos es multiplo de seis ( mod6) ó (°6) mod6 = 130 - 2a °6 = 130 -2a - 132 ; ojo 132 = °6 °6 = -2a -2 => °6 = 2a+2 °6 = 2(a+1) --> (a+1) = °6 Remplazando a = 2^X en (a+1) 2^X + 1 = 6 2^X = 5 ----> X = lg en base 2 de 5 No olvidemos que 2^X ( 2^2X + 1) = 130 = 13×5×2 Observar que es multiplo de 5
Sir, thanks for teaching the 2 methods of a cubic equation. But I think, the student should not waste his time, trying to solve the quadratic equaing the root formula, once he gets numbers like square root of a negative prime number. Very interesting indeed.
set 2^x=s then s^3x + s=130=125 +5=5^3 +5, then s=5 is one answer. (s-5) (s^2 +5s +26)=0. other two answers are composite. consequently x=log(5, 2). Fin
I know u can use first order derivatives to answer these problems. An engineer showed me at grad school; it's like math acrobatics in simplicity. But that is only one shot deal since I don't use calculus at work & school was so long ago, so I've forgotten about it☹️
If you don't have access to Log tables, then... 2^x = 5 So, x is between 2 and 3. When x is 2, value is 4; when x is 3, value is 8; So, if it has to be 5, then the value of x will be x = 2+[(3-2)/(8-5)] = 2.33
Tôi tính nhầm cơ bản ban đầu x€(2÷3). Vì 130 gần 68 hơn 520 nên nó rơi vào đâu đó tầm (2.3 ÷ 2.4) bằng phương pháp nội suy tuyến tính. Tôi chỉ mất 30 giây để khoanh vùng đáp án. Để chính xác thì mất nhiều thời gian hơn. Toán học làm tôi nhức đầu khi các hằng đẳng thức bắt đầu hiện ra
The variables "x" in the equation were exponents not factors though. Although I agree he took the long route, you didn't even understand what he was looking for.
Let’s take x value as 1. You will get 10, let’s take x value as 2 you will get 68, if we take x value as 3 we will get number larger than 130 for sure. So value of x will be between 2 and 3 . Solved.
This is a very long procedure, no need to use logs I can solve this in 5 steps From the first statement remove 2 power x common And the solution solves easily
Да ну! Так уж и понятно прямо "всё" и "всем". Гляди мой коммент, я по-хипстерски решил за минуту, рассписал за две 😂 ...а вот тVой тарабарский "яZiк" здесь точно 99% непонятен 😊
For those who are looking for simple solution. Let’s take x value as 1. You will get 10, let’s take x value as 2 you will get 68, if we take x value as 3 we will get number larger than 130 for sure. So value of x will be between 2 and 3 . Solved.
Simple method would be take logarithm on both sides. It will be a simple equation, i. Ee., K (X) = a number. (N). X=N÷K =Z. Anti log (X)=Anti log (Z). As simple as that. 😄
có thể vào phòng thi bạn phải làm vậy để ra đáp án chính xác. Nhưng thực tế tôi chỉ cần một cái máy tính bỏ túi casio với phương pháp nội suy tôi mất khoảng 20 giây là có đáp án
let t = 2^x t³ + t = 5³ + 5 t(t² + 1) = 5(5² + 1) ...well at this stage it should be obvious that t = 5 so 2^x = 5 log(2^x) = log(5) x * log(2) = log(5) x = log(5)/log(2) ...how did you manage to bloat this to more than 18 minutes ?!! 😂
Ни слова не понял, просто заглянул, что у него там в конце. Я конечно не профессор математики, но по моему у него икс равен 2 умножить на 33, что является бредом. Или у них там точка не умножение означает?
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I used to get straight A’s in math back in college. I am here 30 years later to tell you that none of this shit applies in real life. 😂 😂
8^x+2^x=(2^x)^3+2^x=Y^3+Y=Y(Y^2 +1), with Y=2^x. On the other hand, 130=2*5*13=5*26=5*(5^2 +1). Therefore, Y(Y^2 +1)=5(5^2 +1)Y=5. Therefore 2^x=5 => x=log2(5).
Надо показать, что других действительных корней нет: (Y^2+Y-130):(Y-5)=Y^2+5Y+26. D=25-104=-79
2^x=5 => x=log2(5) by definition;
and also
x ≈ 2,32 (not x=2.33)
That's the approximate value dear
@@pappujha641, and the mathematical symbol of approximate value is ≈ not =, dear
@@evic1025 oh ! I don't knew that
It released my mind from pressure. Thank you so much for solving this.
Very good I am Indian.
Si ; 130 = 13×5×2
Entonces (8^X + 2^X) es multuplo de
13; 5 ó 2
Haciendo : 2^X = a
8^X + 2^X = 130 restando 2a
a^3 + a - 2a = 130 - 2a
a^3 - a = 130 - 2a
a(a^2 - 1) = 130 - 2a
a (a-1)(a+1) = (a-1)(a)(a+1) = 130 -2a
y como todos sabemos... el producto de
tres numeros consecutivos es multiplo de seis ( mod6) ó (°6)
mod6 = 130 - 2a
°6 = 130 -2a - 132 ; ojo 132 = °6
°6 = -2a -2 => °6 = 2a+2
°6 = 2(a+1) --> (a+1) = °6
Remplazando a = 2^X en (a+1)
2^X + 1 = 6
2^X = 5 ----> X = lg en base 2 de 5
No olvidemos que
2^X ( 2^2X + 1) = 130 = 13×5×2
Observar que es multiplo de 5
数学は厳密な値を求めるのが目的だから対数のような
無理数を小数に直して概算で表すのは不適切。正解は
底を2、真数を5とする対数になる。
せめて、小数に直したところで”≒”にしてほしかったですね。
In this case, a math Olympiad candidate would not have wasted time solving the quadratic function.
Sir, thanks for teaching the 2 methods of a cubic equation. But I think, the student should not waste his time, trying to solve the quadratic equaing the root formula, once he gets numbers like square root of a negative prime number. Very interesting indeed.
logc(a) / logc(b) = logb(a) ---> logc5 / logc2 = log2(5) = 2.322
You can use log first itself
Or you can use synthetic division
set 2^x=s then s^3x + s=130=125 +5=5^3 +5, then s=5 is one answer. (s-5) (s^2 +5s +26)=0. other two answers are composite.
consequently x=log(5, 2). Fin
I know u can use first order derivatives to answer these problems. An engineer showed me at grad school; it's like math acrobatics in simplicity. But that is only one shot deal since I don't use calculus at work & school was so long ago, so I've forgotten about it☹️
Expanding the equation... 😊
If you don't have access to Log tables, then...
2^x = 5
So, x is between 2 and 3.
When x is 2, value is 4; when x is 3, value is 8;
So, if it has to be 5, then the value of x will be
x = 2+[(3-2)/(8-5)] = 2.33
2.322 is a closer answer 🎉🎉🎉
Not to be picky, really enjoyed it, shouldn't you use appropriate sign instead of equal?
I completely understand
Это же надо столько времени извлекать корень из отрицательного числа! Я уже подумала, что извлечёт.
Большое спасибо!❤❤❤
In the beginning I could see there is no real solution.
❤ math
Thank God the equation can be factored. I'd hate to have to use the general solution for a cubic equation.
8x8x2+2=130
Well if we substitute the value of x in the above equation it should satisfy the equation will it satisfy
130~128=2^7 2^8=256 2^7 настолько приближено к 130 , что если пренебречь этой разницей , то получим уравнения 2^3х+2^х=2^7. 3х+х=7 х
Pretty problem 😊
Ty, watching it for fun. I miss math ❤
nice math
Thank you for your valuable comment
Amazing
what a mind game
Very nice explaination
Thank you for your valuable comment
If you intend to reject complex roots, you can check the discriminant immediately, without going through the entire quadratic equation.
Seems To He Has Compounded The Problem Well Beyond It's Original Challenge . I'm Not A Math Whiz But I Got x = 13 Long Time Ago Using Way Less Paper .
💯
Bài Toán cũng khó nhưng khá hay nhưng Tôi cũng rất thích học môn Toán và Tôi cảm ơn Thầy...💫🌡️
The scenic route! This is what turns people off of mathematics.
Too easy... Can't be olympiad, 20 secs tops to solve this
It reduces to y^3+y=130
So y=5,
x= log5 base 2
Dho else skipped to the end to find out what the hell X equalled?
Tôi tính nhầm cơ bản ban đầu x€(2÷3). Vì 130 gần 68 hơn 520 nên nó rơi vào đâu đó tầm (2.3 ÷ 2.4) bằng phương pháp nội suy tuyến tính. Tôi chỉ mất 30 giây để khoanh vùng đáp án. Để chính xác thì mất nhiều thời gian hơn. Toán học làm tôi nhức đầu khi các hằng đẳng thức bắt đầu hiện ra
Интересненькое разложение на множители уравненьица!
x = log2(5) The "2" is a subscript.
Don't understand why he rejected complex y at 9:05 and didn't find x as a complex power of 2.
He was solving for the roots, and did not have a method to deal with rhe complex logarithm. There are three roots and two are complex.
@@davidrush4908незачем тогда и это видео снимать было
Ещё и растянул кучу действий, которые можно было прооешать в уме. А он тратил по полчаса на них
You´re a great teacher. Thanks for sharing your knowledge.
Thank you too for your valuable comment
Что конкретно можно вычислить этим примером на практике?
человека которй не умеет в математику
Долго ему пришлось возиться, чтобы понять, что уравнение не имеет корней и решения. Однако всë же интересно.
8x +2x = 10x 10x= 130 x= 13 Easy peasy
8x= 104, 2x= 26, 104+26= 130
All done in one minute, none of the "BS" you just went through
The variables "x" in the equation were exponents not factors though. Although I agree he took the long route, you didn't even understand what he was looking for.
Let’s take x value as 1. You will get 10, let’s take x value as 2 you will get 68, if we take x value as 3 we will get number larger than 130 for sure. So value of x will be between 2 and 3 . Solved.
How many felt frustrated at the pace at which the problem was solved? It was like talking to flash in zootopia. 😂
Please use simples methods of solving the problem,don't complicated.
Who the hell uses this in present life.
😂
5 cubed minus 5 equals to 120. 2 cubed, which was y cubed, is 8, plus y, which is 2, equals 130.
❤❤❤
Solution by insight
2^3x +2^x=130
2^x(2^2x+1)=130
5(5^2+1)=130
2^x=5
x=log_2 (5)
This is a very long procedure, no need to use logs
I can solve this in 5 steps
From the first statement remove 2 power x common
And the solution solves easily
Fiz de cálculo mental em 10 segundos. Não cansei a cabeça e não gastei tinta da caneta. Foi por isso que nunca gostei de matemática.
What are u doing now?
Making it too complex
Ну зачем так подробно!!! Все же понятно и так.
Да ну! Так уж и понятно прямо "всё" и "всем". Гляди мой коммент, я по-хипстерски решил за минуту, рассписал за две 😂
...а вот тVой тарабарский "яZiк" здесь точно 99% непонятен 😊
@@b213videoz Ну так хрюкай на свиномове, чего мучаешься на тарабарском.
Krieg ich Kopfschmerzen. Unfassbar.
13
До 2мин молодца. Угерек в 3степени плюс угерек равен 5в третьей степени плюс 5 игерек равен 5, икс равен корень квадратный из 5.
8^x+2^x=130 => 8^x=130-(2^x), so therefore, I cannot answer ur question :(
2.6
I think you have to know y=5 before you start separating 130 into 125+5. Way too much extra work.
At 2:24 it was completely obvious that y=5. Everything from there to the x=log2(5) step is absolutely unnecessary.
تمرين جميل جيد . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين .
He has made the video so to gain the money
Too easy
簡單的問題，複雜的解法😂😂
도움이 되었습니다
8*13 +2*13 =130
Good explanation to understand ! Thax.
Опять отвратительная концовка с приближенным вычислении. Не надо приближенное!
8×11 + 2×11 = 130
× = 11
A Nice Olympiad Exponential Problem: 8^x + 2^x = 130; x = ?
8^x + 2^x = 130; 130 > 8^x > 2^x > 0
First method:
8^x + 2^x = 130 = (5)(26) = (5)(25 + 1) = 5^3 + 5^1
Convert the exponential base number 5 into 2 using logarithmic math:
Let: 2^n = 5, n = log₂5 = 2.322; 5 = 2^2.322
8^x + 2^x = 5^3 + 5^1 = (2^2.322)^3 + (2^2.322)^1 = 8^2.322 + 2^2.322; x = 2.322
Second method:
8^x + 2^x - 130 = [(2^x)^3 - 5^3] + (2^x - 5) = (2^x - 5)[(2^x)^2 + 5(2^x) + 26] = 0
2^x - 5 = 0 or (2^x)^2 + 5(2^x) + 26 = 0
2^x = 5, x = log₂5 = 2.322 or 2^x = (- 5 ± i√79)/2; 2^x > 0 is not defined
Complex roots are meaningless for the exponential variable 2^x; Rejected
Answer check:
As shown in First method; Confirmed
Final answer:
x = log₂5 = 2.322
This answer it's currect bro well done...
For those who are looking for simple solution. Let’s take x value as 1. You will get 10, let’s take x value as 2 you will get 68, if we take x value as 3 we will get number larger than 130 for sure. So value of x will be between 2 and 3 . Solved.
Заумно как-то , по моему можно проще.
प्लास, भेलू😂😮😅😊
Troche zawile ale wyszlo mi tak samo..
I’m more stuck on whether it’s y or ooowuheye
3
Simple method would be take logarithm on both sides.
It will be a simple equation, i. Ee.,
K (X) = a number. (N).
X=N÷K =Z.
Anti log (X)=Anti log (Z).
As simple as that. 😄
Answer no right‼️‼️‼️‼️
CHATGPT ANSWER = x ≈ 2.3092 (rounded to 4 decimal places).
Js do guess and check 8*13+2*13=130
i like the pawar😆
Whew! it just went above my head....
Why do you have to remind those school days...
Just chill and let others enjoy others life...
Joke..
Eu sabia fazer esse cálculo com maçãs 😂
This guy litteraly Torturing himself to solve this problem bruh
có thể vào phòng thi bạn phải làm vậy để ra đáp án chính xác. Nhưng thực tế tôi chỉ cần một cái máy tính bỏ túi casio với phương pháp nội suy tôi mất khoảng 20 giây là có đáp án
😢😢😢😢😢ฝ
😮😮😮😅😮😮😮😮😮
It was already solved literally like by second minute he took soooo long way to end 😂
Ghetto language now in math? Trashy.
Yes bro. Very critical minded person.
ITNA KARNE KE BAAD BHI APPROX HI NIKAAL PAA RHE SINGLE VARIABLE ITNA RULA DIYA DOUBLE VARIABLE HOTE TO LAST MAI SUICIDE LETTER LIKHNA PAD JAATA
Without mathicks man is consider Hichadda
let t = 2^x
t³ + t = 5³ + 5
t(t² + 1) = 5(5² + 1)
...well at this stage it should be obvious that t = 5
so 2^x = 5
log(2^x) = log(5)
x * log(2) = log(5)
x = log(5)/log(2)
...how did you manage to bloat this to more than 18 minutes ?!! 😂
thank god for that, Now i can sleep at nights
Yeah, we all didnt sleep all this 50 yrs.😅
А так все просто начиналось😂😂
Ни слова не понял, просто заглянул, что у него там в конце. Я конечно не профессор математики, но по моему у него икс равен 2 умножить на 33, что является бредом. Или у них там точка не умножение означает?
(A±B)^2=A^2±2AB+B^2
Credo che tu abbia sbagliato il quadrato di binomio: y^2+10y+25 😊
Accurate answer is =2.322, sir.
2,321928, to be more accurate
X= 13
Ответ можно получить проще:
X=(ln130-ln9)/ln2.
Twice the same calculation… what a torture…
And go to imaginary numbers just to reject them, there is something ridiculous…
Não gosto de Álgebra,para que serve ?
🤯🤯🤯
y € (2 :3).