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
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.
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
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.
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.
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
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
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☹️
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 😊
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. 😄
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
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 ?!! 😂
Ugh... seriously 13 boys and girls. Without the nonsense. ❤ took 6 seconds. THINK. 😊 When teachers like this over Demo. Is reason why American schools are a major issue. P.s. I live in America
19 minutes to solve an equation requesting a max of 5 minutes to a student with an average IQ … just a terrible waste of time to get views on UA-cam… and a terrible headache for every math teacher watching by curiosity
F y 8^x+2^x=130 (2^x)³+2^x=130 t³+t-130=0 1. 0. 1. -130 5. 5. 25. 130 1. 5. 26. 0 (t-5)(t²+5t+26)=0 t-5=0 t=5 t²+5t+26=0 This solution is rejected because it is not a real number. So, 2^x=5 log2^x=log5 xlog2=log5 x=log5/log2 x=logbase2of5.
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
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. 😂 😂
2^x=5 => x=log2(5) by definition;
and also
x ≈ 2,32 (not x=2.33)
That's the approximate value dear
@@Pappujha710, and the mathematical symbol of approximate value is ≈ not =, dear
@@evic1025 oh ! I don't knew that
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
logc(a) / logc(b) = logb(a) ---> logc5 / logc2 = log2(5) = 2.322
It released my mind from pressure. Thank you so much for solving this.
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.
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
2.322 is a closer answer 🎉🎉🎉
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。
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.
Very good I am Indian.
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.
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
Это же надо столько времени извлекать корень из отрицательного числа! Я уже подумала, что извлечёт.
Expanding the equation... 😊
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
13
You can use log first itself
Or you can use synthetic division
In this case, a math Olympiad candidate would not have wasted time solving the quadratic function.
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.
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☹️
You´re a great teacher. Thanks for sharing your knowledge.
Thank you too for your valuable comment
8x8x2+2=130
I completely understand
❤ math
In the beginning I could see there is no real solution.
If you intend to reject complex roots, you can check the discriminant immediately, without going through the entire quadratic equation.
8*13 +2*13 =130
تمرين جميل جيد . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين .
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)
Что конкретно можно вычислить этим примером на практике?
человека которй не умеет в математику
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
Unnecessary prolongation off solution after obvious answer i.e. 5.
what a mind game
Not to be picky, really enjoyed it, shouldn't you use appropriate sign instead of equal?
До 2мин молодца. Угерек в 3степени плюс угерек равен 5в третьей степени плюс 5 игерек равен 5, икс равен корень квадратный из 5.
Ty, watching it for fun. I miss math ❤
What are u doing now?
Интересненькое разложение на множители уравненьица!
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 .
At 2:24 it was completely obvious that y=5. Everything from there to the x=log2(5) step is absolutely unnecessary.
x = log2(5) The "2" is a subscript.
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
Долго ему пришлось возиться, чтобы понять, что уравнение не имеет корней и решения. Однако всë же интересно.
Опять отвратительная концовка с приближенным вычислении. Не надо приближенное!
Amazing
The scenic route! This is what turns people off of mathematics.
Well if we substitute the value of x in the above equation it should satisfy the equation will it satisfy
Thank God the equation can be factored. I'd hate to have to use the general solution for a cubic equation.
nice math
Thank you for your valuable comment
5 cubed minus 5 equals to 120. 2 cubed, which was y cubed, is 8, plus y, which is 2, equals 130.
Большое спасибо!❤❤❤
Please use simples methods of solving the problem,don't complicated.
Who the hell uses this in present life.
😂
12
3
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
Krieg ich Kopfschmerzen. Unfassbar.
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.
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. 😄
Good explanation to understand ! Thax.
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
CHATGPT ANSWER = x ≈ 2.3092 (rounded to 4 decimal places).
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 ?!! 😂
2.6
I think you have to know y=5 before you start separating 130 into 125+5. Way too much extra work.
（2^3）^x+2^x=130
(2^x)^3-5^3+2^x-5=0
(2^x-5)[2^2x+5(2^x)+
25)+(2^x-5)=0
(2^x-5)[2^2x+5(2^x)
+26]=0
2^x-5=0👈
2^x=5
Or
2^2x+5(2^x)+26=0👈
2^x={-5+/-[(5^2)-
4(1)(26)]^1/2]}2(rej.)
Delta
Pretty problem 😊
Accurate answer is =2.322, sir.
2,321928, to be more accurate
Very nice explaination
Thank you for your valuable comment
Ugh... seriously 13 boys and girls.
Without the nonsense. ❤ took 6 seconds.
THINK. 😊
When teachers like this over
Demo. Is reason why American schools are a major issue.
P.s. I live in America
Dho else skipped to the end to find out what the hell X equalled?
Making it too complex
(A±B)^2=A^2±2AB+B^2
Credo che tu abbia sbagliato il quadrato di binomio: y^2+10y+25 😊
X =13
Js do guess and check 8*13+2*13=130
8^x+2^x=130 => 8^x=130-(2^x), so therefore, I cannot answer ur question :(
2^x*(2^x)^2 +2^x=130
m=2^x
m^(3)+m-130=0
Adivinar: m=5
m^(3)+m-130:(m-5)=m^(2)+5m+26=0
Resultados no reales
Por eso: m=5
x*ln(2)=ln(5)
x=ln(5)/ln(2)
x = 2,322
28 × 3 + 63 × 4 = 336
Bài Toán cũng khó nhưng khá hay nhưng Tôi cũng rất thích học môn Toán và Tôi cảm ơn Thầy...💫🌡️
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.
Too easy
X= 8
He has made the video so to gain the money
(y)3 -(5)3 +y -5 =0; y=5
💯
19 minutes to solve an equation requesting a max of 5 minutes to a student with an average IQ … just a terrible waste of time to get views on UA-cam… and a terrible headache for every math teacher watching by curiosity
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
Заумно как-то , по моему можно проще.
प्लास, भेलू😂😮😅😊
❤❤❤
8×11 + 2×11 = 130
× = 11
F y
8^x+2^x=130
(2^x)³+2^x=130
t³+t-130=0
1. 0. 1. -130
5. 5. 25. 130
1. 5. 26. 0
(t-5)(t²+5t+26)=0
t-5=0
t=5
t²+5t+26=0
This solution is rejected because it is not a real number.
So, 2^x=5
log2^x=log5
xlog2=log5
x=log5/log2
x=logbase2of5.
B= D
Answer no right‼️‼️‼️‼️
Ответ можно получить проще:
X=(ln130-ln9)/ln2.
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..
Without mathicks man is consider Hichadda
These kinds of pointless things are educated at schools worldwide. What a waste of our youth. You never ever going to need or apply this knowledge.
🤯🤯🤯