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.
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.
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.
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.
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 😊)
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
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?
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.
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.
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.
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.
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.
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.
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...
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).
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.
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.
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.
@@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.
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)
@@marianneoelund2940@marianneoelund2940 Thanks, I understand that, but no other (y-3) on the left is factored/cancelled. I get the 9+ 1 afterwards but.......
@@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.
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!
@@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.
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
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.
i would of been out of paper after that equation. Unless you are a scientist developing theories of flux capacitors majority of people will not need this kind of math.
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.
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.
@@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.
You might be driving people nuts with this very long choice of path you've chosen. Instead of simplicity you chose complexity when math itself is a fearsome subject. How would ordinary mortals get encouraged?
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.
6
The answer is 3
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.
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.
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.
@@oahuhawaii2141Yes. The natural logarithm has to be taken, of course.
참 힘들게 푸시네
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.
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 😊)
That pen writes quite beautifully
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
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?
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.
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.
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.
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.
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.
You could apply the log formula right at the start.
log(x + y) = ?
let y=2^x
then y^3 + y = 3^3 + 3
=> y=3
=> x = log3/log2
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.
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...
@@praiseYAHalways wrong. 8 cube and 8x 3 is different
One in top righ of eight, another one in top right of number two. I found two of them! Genius!
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).
Knowing the theorems was essential to solving for X.
Can we not directly apply log to the main equation and solve
But how can you get a 14 minutes video?
Simple answer: no. You'd still have to solve log(8^x+2^x) = log(30)
You are a genius😊by the why I'm estifanos' son. I'm going n grade 5 and I enjoy your math videos.😊
Right off the bat, x
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).
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.
Well now I know why I just did applied maths, not pure !!, well done
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.
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.
@@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.
Like you can solve in 5 seconds in your head ...
Or 10 pages of math
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
I'm sorry, but the answer is obviously "pizza"
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)
Symbolab gives your result. Butis it a matter of separating exponents?
Is there an order of operations I'm not seeing?
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
Also a²+b²≠(a+b)²
Тот случай, когда ответ, состоящий из логарифмов, оказался сложнее, чем изначальный пример.
At 4.28 where is 1 come from? I lost it there.
Ok nevermind after watching another video I understand it now
Where? I am confused
I lost it there too-whats the explanation for the 1, or would be grateful for the other video URL please?@@wizard105able
@@cornerboytepWhen he converts y-3 into a 1
How about
8 Sq by 2, 2 Sq by 7.
16 +14 =30
X=1.585 by inspection without writing anything, a couple of approximations were enough in about 3 minutes. Anyway congrats for all your effort.
x = log₂(3) = log(3)/log(2) ≈ 1.5849625007211561814537389439478
Note the "approximately equal to" symbol vs. the "equal to" symbol.
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.
Thats because (-3 +- sqrt(31)i) / 2 Cannot be solved, thus it can't be simplified.
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
The result equation contains only constants, thus it is final, even if it appears messy.
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
Y are you pointing at the rabbit? Because we are now going to follow it all the way down!
(8 +2)*=30 10 *=30 30:10=3
Как я любила их решать в школе!
เก่งมาก😊
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😂
After grading, you'll find another one covering your answer sheet.
Soruyu çözerken onluk log yerine ln de kullanılabilirdin daha şık estetik bir çözüm olurdu
So, what is log 2(3) ????
Can't we have a straight answer in numbers??? 😋
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. 😅
Вручную, подобрал сразу. Чего вола катает...
Тоже хотел написать. Волосы дыбом от таких "полезных" манипуляций.
@@tz8811 сразу видно что 3=x про логорифмы уже не помню, с этим перегнул!)
2^x=3, x=log 3/log 2.
(8X+2X)=10x=30...-> /÷10..-> x=3
Why wouldnt it be: x= log10(30)? which is x=30
Wonderfully explained
Hm. I think after substituting y, splitting 30 into 27+3 will only occur to you if you already know the solution.
Not really, actually that simple trick is a tool you use lot in calculus
Nice trick
Make in math octal
a=2^x, a^3+a=30, a(a^2 + 1)=30=3*10, a=3, a=2^x =3, x=log_2 3
Gets harder as we go
I have major doubt atleast 8 square chesina 64:
64+some number=30
How is it possible
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.
Yes there is another way
I get it except for how (y-3) became 1 at 4:27
(y - 3) is a common multiplier and is being factored out to the left. Reverse distributive law.
@@marianneoelund2940@marianneoelund2940 Thanks, I understand that, but no other (y-3) on the left is factored/cancelled. I get the 9+ 1 afterwards but.......
@@carlw It's straightforward distributive law:
(y-3)*z + (y-3) = (y-3)*(z+1)
Just factor out (y-3). Nothing is being canceled.
@@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.
Esuatt start oddition.after result free Master combine
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!
What? Is not the same
Simplification??
X equals 3, did it it 15 seconds. Love being a math geek :)
I think you meant 2 raised to the power X is 3.
But you missed the complex valued solutions.
I looked at the first page and intuitively made the same conclusions as @twinmomsurvival5744 can you explain how your answer is correct?
@@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.
Nope.
@@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?
Better than ASMR
Very annoying music, please change it. I wanted to quit watching right in the middle of the solution
Mute it .-.
@@joelo555 If he couldn't figure that out, he probably can't follow the math. 🤔
I advise you to turn the sound off.
Pretty simple.
Right please remake silent 🔇
Agree. Music is dreadful.
y^3+y=30 3 is ok so (y-3)*Q=0 and then easy to the end
What?😊
thanks :)
X = logaritmo de 3 en base 2
Its a nice sweetch lieder number for churche master.29+(9)XYC=23(8)+2(/29)__3#CH>133IPS=122+(N14)+S128
That's the real solution, but there are also complex solutions
Yes, an infinite number of them.
3x 8 = 24, 3x 2 = 6. 6+24 = 30
It's very complicated maths can't get my head around.
Doesn't follow any particular rule.
I got a 99 on my algebra regions but couldn't add anything today. 😅
Меня всегда интересовал вопрос : "а нахер это всё надо?:
Километр бумаги истратили
Well you didn't get the -2 for not showing all your work.
Did I see a word 'simplification' , yet the final answer is ' not possible ' ....... Oh me, oh my... How dumb I am
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
Is that supposed to be a 'simplification'? How does 'y' come into an equation that has no 'y' to start with?
The y is only a symbol, just like x. It’s easier to work with one variable (y) than to work with 2^x.
@@robbie330 what is 2 ^?
The judges say yes the answer is 3 they roll big joints too
Complicated version. Surely you don't expect to make math interesting with your extended method?
It's simple. Take the x's out, and you get 10.
I have a solution for cake failing to rise.
3
Rien que la deuxième ligne...?
C'est quoi tout ces a, m, etc....?
8x3 + 2x3=30 ( value X is 3 )
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.
@@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
Dumb
Did it faster in my head without all the extra work.
It took me 5 sec in the head
That's a claim you will never be able to prove,
Sir pen ka naam Bata do plz ❤❤
i would of been out of paper after that equation. Unless you are a scientist developing theories of flux capacitors majority of people will not need this kind of math.
👍👍👍
Only 1 munite is enough for this example 😊
2^x=3
Log2(3)
Why do you show the problem, right “Solution” underlined, and then rewrite the problem? Seems redundant and a waste of time.
Я не математик но (8*3 = 24) + (2*3 = 6) = 30, я не понял огромной формулы но ответ верно высчитал
Это студенческий пример, ты не вытянешь. Ты не отличаешь степень от произведения. И "x" равен не "три", а вот этому 7:03
Логарифм называется.
Very good
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.
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.
But it's still 3
@@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.
x = 1.5849625
What tha hell??
Muito bom
I found two X’s. One next to the 8 and the other next to the 2..
After grading, you'll have another one that's easier to find -- it covers your whole answer sheet.
tease with cross ... 🤞 😅
Off course its the long way. Thr algorithm pays by the hour
What?
It's right there above 8 and 2.
They raise those bases to the power of x.
(8+2)x = 30
10x = 30
X= 3
🤔🤔🤔🤔
Dumb
You might be driving people nuts with this very long choice of path you've chosen. Instead of simplicity you chose complexity when math itself is a fearsome subject. How would ordinary mortals get encouraged?
Why make life hard. Let x=2 then it is obvious that x must have a fractional value somewhere between 1 and 2.
When I asked Chatgpt, x is about 1.65.
2
please. do quadratic (x-7) =3, tks
Where do you want the square? It's missing from your expression.
Bro lost me when he turned 8x into 23