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
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"
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.
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.
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.
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!
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
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
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!
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)].
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?
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.
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
x = 2 is not a solution, x = 1 is the only real solution.
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"
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.
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.
x=1 is a solution
X = -1 is not a solution.
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.
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.
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.
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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!
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
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
x = 1
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)
hocam ya yine karıştırdın. Gitti güzelim soru
X =1, only real sol
Substitution did you wrong
Your answer is completely wrong.
The correct answer is x=1.
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!
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)].
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?
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)
By AM GM inequality a=b=c=d then you can put this in the second equation and get answer