As my math teacher used to say: "You can call the dog 'cat' just don't forget about it." Apparently you forgot about it, lol.
@eager4835
Жыл бұрын
Put 2^x=t, then numerator becomes t^4+t^2+1, we can factor it by add and subtract t^2. So it become (t^2+t+1)(t^2-t+1). We can cancel t^2+t+1 from numerator and denominator and we left with t^2-t+1=3 which is simple quadratic
@goldfing5898
Жыл бұрын
4:00 If you look St the quartic t^4 - 2t^2 - 3t - 2 = 0 or t^4 = 2t^2 + 3t +2, You can, besides of t = -1, also easily spot t = 2 ad the second solution, since 2^4 = 16 and 2*2^2 + 3*2 + 2 = 2*4 + 6 + 2 = 8 + 8 = 16, too.
@goldfing5898
Жыл бұрын
Then multiply the corwspomding linear factors: (t + 1)(t - 2) = t^2 - t - 2 and divide (t^4 - 2t^2 - 3t - 2) : (t^2 - t - 2) = t^2 + t + 1 = 0 and this remaining quadratic has only complex solutions.
@goldfing5898
Жыл бұрын
For the second method, why do you write x rather than t? After all, you must substitute t = 2^x. In the end, we get t = 2 = 2^x, thus x = 1 as the only real solution.
@anad8341
10 ай бұрын
I was going to suggest using long division, but factoring the numberator is also a good idea. The fist method really doesn't help much.
@vaggelissmyrniotis2194
Жыл бұрын
Other two group of solutions are x=2π*i*(k+2/3)/ln2 and x=(k+1/3)*2π*i/ln2,k be an integer.
@dariosilva85
Жыл бұрын
x = 2 is not a solution, x = 1 is the only real solution.
@armacham
Жыл бұрын
7:25 He wrote the problem wrong, he replaced 2^x with x, 4^x with x^2, etc. He would have been fine if he just replaced 2^x with something else, like "t"
@Tomorrow32
Жыл бұрын
X = -1 is not a solution.
@fabiotiburzi
Жыл бұрын
x=1 is a solution
@sy8146
Жыл бұрын
Thank you for explaining. [1] In this video, concerning 2nd method, using x and t is a little strange, actually "x=2=2^x, and x=2, x=-1" is confusing expressing. t = 2^x = 2, -1 is better expression. [2] And, as for method 1, after getting (t+1)(t^3-t^2-t-2)=0, we get (t+1)(t-2)(t^2+t+1)=0. For t^2+t+1=0, Δ=-3
@mathbunroeung7019
Жыл бұрын
❤❤❤
@scottleung9587
Жыл бұрын
I think the real solution you're looking for is x=1, which should be obvious from the start. However, I went for gold and found the other three solutions, which come in the form of families. Two of these three families turn out to be real solutions, and only the last one is complex. Although this is up for debate, I'm not sure if every single step I took was correct!
@seanfraser3125
Жыл бұрын
Use substitution u = 2^x: (u^4+u^2+1)/(u^2+u+1) = 3 Multiply both sides by the denominator and move everything to the left hand side: u^4 - 2u^2 - 3u - 2 = 0 Through a bit of guessing and testing, I found u=-1 and u=2 to be solutions to the above quartic. Using this we can factor: (u+1)(u-2)(u^2+u+1) = 0 Since the quadratic factor has negative discriminant, we have no more real solutions. There is no real x such that 2^x = -1. If u=2, then x=1. Thus our only real solution is x=1
@itzerr5547
Жыл бұрын
I have made system of eqn, please solve this for real solutions : (a+b+c+d)/4 = 4th root of (abcd).........(1) ln(ln(ln(a^b^c^2))) = 2........(2)
@rakenzarnsworld2
Жыл бұрын
x = 1
@ahmetsen5601
Жыл бұрын
hocam ya yine karıştırdın. Gitti güzelim soru
@kianmath71
Жыл бұрын
X =1, only real sol
@knutthompson7879
Жыл бұрын
Substitution did you wrong
@mimathematics287
Жыл бұрын
Your answer is completely wrong. The correct answer is x=1.
@logminusone1272
Жыл бұрын
I feel the solution can be arrived in a far simpler way. Use the formula (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca to simplify the numerator. The numerator can be written as: (16^x + 4^x + 1) = (4^x + 2^x + 1)^2 - 2.2^x(4^x + 2^x + 1) Now, divide by the denominator (4^x + 2^x + 1). You are left with 4^x - 2^x + 1 = 3. Now, put p = 2^x. You get, p^2 - p - 2 = 0 or (p + 1)(p - 2) = 0. Then on, it is simple. Either p = 2^x = -1, or p = 2^x = 2. The first case yields x = i*log2 (base e). The second yields x = 1. Take your pick of the solution!
@forcelifeforce
Жыл бұрын
2.2 means 2 + 2/10. It is not 2 times 2. And, 2.2^x means (2 plus 2/10) raised to the x power. Here are a couple of redos: The numerator can be written as: (16^x + 4^x + 1) = (4^x + 2^x + 1)^2 - 2*2^x(4^x + 2^x + 1) *or* (16^x + 4^x + 1) = (4^x + 2^x + 1)^2 - 2[2^x(4^x + 2^x + 1)].
@logminusone1272
Жыл бұрын
The . (dot) notation to denote multiplication was popular when we were in school working on long algebraic expressions. Maybe it is not popular now. Anyway, the idea is understood, as I can see. While writing or typing, the "x" multiplication sign may often get confused with a variable x. But yes, both "x" and "." notations have their own weaknesses as they can both get confused for something else. So "*" is good. But do we use "*" while writing?
@lordofdarkness4232
Жыл бұрын
I have made system of eqn, please solve this for real solutions : (a+b+c+d)/4 = 4th root of (abcd).........(1) ln(ln(ln(a^b^c^2))) = 2........(2)
@marianne-wt8it
Жыл бұрын
By AM GM inequality a=b=c=d then you can put this in the second equation and get answer
Пікірлер: 27