Given: x↑x = 1/x To find: x Rewriting 1/x as x↑(-1): x↑x = x↑(-1) Taking log of both sides: log(x↑x) = log(x↑(-1)) As log(a↑b) = b·log(a): x·log(x) = -log(x) Bringing all terms to LHS: x·log(x) + log(x) = 0 log(x)·(x + 1) = 0 From here, log(x) = 0, or x = -1 log(x) = 0 → x = 1. Therefore, x = ±1
Here is my solution. It is similar to your "first" method, but using the absolute value, as you suggested, and with the solutions to the equation aᵇ=aᶜ justified. As 1/x=x⁻¹, our equation is equivalent to xˣ=x⁻¹. This equation is of the form aᵇ=aᶜ. The solutions to this equation are: a=0 and b=c=0 or a=0 and b>0 and c>0 or b=c or a=1 or a=-1 and aᵇ & aᶜ have the same sign. (proof given below) In our equation, we can't have x=0, so this gives: x=-1 or x=1 or x=-1 and (-1)⁻¹ & (-1)⁻¹ have same sign, which of course they do. So the solutions to our equation are x=-1 and x=1. Proof of the solution of aᵇ=aᶜ. If a=0, we have 3 cases for aˣ, i.e. for 0ˣ: If x0, we have 0ˣ=0 So if a=0, and aᵇ=aᶜ, either aᵇ=1=aᶜ i.e. 0ᵇ=1=0ᶜ⇔b=c=0 or aᵇ=0=aᶜ i.e. 0ᵇ=0=0ᶜ⇔b>0 and c>0. If a≠0, aᵇ=aᶜ ⇔|aᵇ|=|aᶜ| and aᵇ & aᶜ have the same sign (the 2nd condition is implicit in the following lines until it reappears) ⇔|a|ᵇ=|a|ᶜ (using the result that if x and y are real and xʸ is defined as a real number then |xʸ|=|x|ʸ) As a≠0, |a|>0, so we can take ln of both sides So |a|ᵇ=|a|ᶜ ⇔ln|a|ᵇ=ln|a|ᶜ ⇔b ln|a|=c ln|a| ⇔(b-c)ln|a|=0 ⇔b-c=0 or ln|a|=0 (and aᵇ & aᶜ have same sign) ⇔b=c or (|a|=1 and aᵇ & aᶜ have same sign) ⇔b=c or a=1 or (a=-1 and aᵇ & aᶜ have same sign) For a proof of the result that if x and y are real and xʸ is defined as a real number then |xʸ|=|x|ʸ see my comment starting "Nice question, as usual. Here's my analytical solution for real x." on SyberMath's video "Solving a Super Exponential Equation" ruclips.net/video/j_uS15_hAxk/видео.html .
Sir (e to the power y)to the power(e to the power y) = e to the power (y*(e to the power y)) = e to the power -y therefore e to the power y = -1 implies x = -1 However there is an exception that 1 to the power 1 equals to 1／1
@@SyberMath Your method with ln is false: (x+1)lnx =0 gives only 1 as solution But ln(-1) not defined. The good method is: x^p=1 , p=x+1 is equivalent to: {• p=0 and x0 : -1 is solution, • or x=1 and p exists: 1 is solution, • or x=-1 and p EVEN: -1 is solution bcse p=-1+1=0 is EVEN } This RULE is in: ruclips.net/video/aGBe9ObqlwA/видео.html
@@Frank-kx4hc I'm sorry, Sir Frank, but zero is neither odd nor even. Remember what zero represents - the lack of value. The number line is discontinuous at zero. If you want to prove something: remember that odd plus odd is even? or, put odd times odd is odd. If you use zero, you will see you can't prove zero to either odd, or even. You see, if you apply number theory you will see the point.
@@Roq-stone without zigzag: • (-1)^0 =1: so x=-1 is solution of your own equation x^(x+1)=1 , • any number (2k+1) is odd for all k in Z, • any number (2k) is even for all k in Z. Thus no need to speak about : x=0, 0 is neither odd nor even, number theory,...
I knew that although x = -1 might have worked, that was extraneous due to the fact of ln(x). Plus I figure to use properties of ln(1/x) being ln1-lnx. This is not a challenging problem at all.
oh. its holly war of roots of equation f(x)^g(x)=f(x)^h(x)
Год назад
It's fascinating how people (most of them certainly smarter than me) avoids understanding what Sybermath is doing here (maybe he wont even agree with me. But I say he's more right than he will admit). Once you have the algebraic expression giving x=-1 as an additional solution. Then that's that. It is obviously correct and -- if you insist -- easily verified. It has nothing to do with ln(-1) not being defined. That might be interesting in itself but irrelevant here. Always trust algebra over some applied mathematics exp or something. The simpler and clearer definitions the more powerful the use of them. Being vague to the point of reaching no results isn't that productive. Even mathematicians need to be a LITTLE productive. Even pure ones.
Quite interesting problem. Start with multiplying x on both sides. x^x * x = 1 Since we have x^x * x^1, we can add those two exponents together. x^(x+1) = 1 The interesting part here is that if we take the x+1th root on both sides we get that x= x+1 root of 1. But because any nth root of 1 is just equal to 1 (for n not equal to zero which grants undefined) we get that x = 1, which indeed is a solution. If we then take log_x on both left and right-hand sides we get that x+1 = log_x(1) (assuming that in this case x is not 0 or 1) we can say that x+1 = 0 leaving us with x=-1 which is another solution. I know this way of solving it is very moot as there are a lot of false assumptions such that logx(1) = 0 even though it isn’t always true. But funnily, it does give the two solutions that we want so… x = 1 or -1
For Y = X^X the graph in negtive domain has (-1, -1) point which is the only intercourse with Y = 1/X So the graph should show the (-1, -1) green point along with the positive domain curve line.
The way Desmos works is before even solving any equation it takes a look at the domain even though negative integers are allowed it will not yield a continuous function so just by convention x^x is defined for all x>0 and that’s why you don’t see a green dot.
No, not quite. Technically, the symbols x^x is an abbreviation for the more correct expression exp(x·ln(x)), which is only defined for x > 0. This is a really bad habit people online have, of using the symbol b^a to mean exp(a·ln(b)), when actually, b^a means something completely different than exp(a·ln(b)), and they have different domains (b^a is defined for all nonzero b and integer a, exp(a·ln(b)) is only defined for b > 0 but all a).
@@angelmendez-rivera351 actually both are defined and real if b is negative and a is an integer. If b < 0 then ln(b) = ln(-b) + i*pi and so exp(a*ln(b)) = exp(a*ln(-b)) * exp(a*i*pi) = +- (-b)^a. Sign depends on if a is even or odd respectively and the expression is real because -b > 0.
@@ngdluu ln(x)=0 - legal: e^(ln(x))=e^0 x=e^0=1 but ln(-1) is not defined on real set. if we use this treak, we can both part of every equation multiply by zero and than got 0=0, so cool, x € R. details not matter.
No it's not valid as it stands (assuming we are working in the real numbers). That's why SyberMath recommended using the absolute value. I suggest you see my comment for a solution using the absolute value.
Your method with ln is false: (x+1)lnx =0 gives only 1 as solution But ln(-1) not defined. The good method is: x^p=1 , p=x+1 is equivalent to: {• p=0 and x0 : -1 is solution, • or x=1 and p exists: 1 is solution, • or x=-1 and p EVEN: -1 is solution bcse p=-1+1=0 is EVEN } This RULE is in: ruclips.net/video/aGBe9ObqlwA/видео.html
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Got 'em both using both methods!
Given:
x↑x = 1/x
To find:
x
Rewriting 1/x as x↑(-1):
x↑x = x↑(-1)
Taking log of both sides:
log(x↑x) = log(x↑(-1))
As log(a↑b) = b·log(a):
x·log(x) = -log(x)
Bringing all terms to LHS:
x·log(x) + log(x) = 0
log(x)·(x + 1) = 0
From here, log(x) = 0, or x = -1
log(x) = 0 → x = 1.
Therefore, x = ±1
Right, but not. By taking the log you assume x>0 so you are wrong in concluding that x = -1.
Here is my solution. It is similar to your "first" method, but using the absolute value, as you suggested, and with the solutions to the equation aᵇ=aᶜ justified.
As 1/x=x⁻¹, our equation is equivalent to
xˣ=x⁻¹.
This equation is of the form aᵇ=aᶜ.
The solutions to this equation are:
a=0 and b=c=0
or a=0 and b>0 and c>0
or b=c
or a=1
or a=-1 and aᵇ & aᶜ have the same sign.
(proof given below)
In our equation, we can't have x=0, so this gives:
x=-1
or x=1
or x=-1 and (-1)⁻¹ & (-1)⁻¹ have same sign, which of course they do.
So the solutions to our equation are x=-1 and x=1.
Proof of the solution of aᵇ=aᶜ.
If a=0, we have 3 cases for aˣ, i.e. for 0ˣ:
If x0, we have 0ˣ=0
So if a=0, and aᵇ=aᶜ,
either aᵇ=1=aᶜ i.e. 0ᵇ=1=0ᶜ⇔b=c=0
or aᵇ=0=aᶜ i.e. 0ᵇ=0=0ᶜ⇔b>0 and c>0.
If a≠0,
aᵇ=aᶜ
⇔|aᵇ|=|aᶜ| and aᵇ & aᶜ have the same sign (the 2nd condition is implicit in the following lines until it reappears)
⇔|a|ᵇ=|a|ᶜ (using the result that if x and y are real and xʸ is defined as a real number then |xʸ|=|x|ʸ)
As a≠0, |a|>0, so we can take ln of both sides
So |a|ᵇ=|a|ᶜ
⇔ln|a|ᵇ=ln|a|ᶜ
⇔b ln|a|=c ln|a|
⇔(b-c)ln|a|=0
⇔b-c=0 or ln|a|=0 (and aᵇ & aᶜ have same sign)
⇔b=c or (|a|=1 and aᵇ & aᶜ have same sign)
⇔b=c or a=1 or (a=-1 and aᵇ & aᶜ have same sign)
For a proof of the result that if x and y are real and xʸ is defined as a real number then |xʸ|=|x|ʸ see my comment starting "Nice question, as usual. Here's my analytical solution for real x." on SyberMath's video "Solving a Super Exponential Equation" ruclips.net/video/j_uS15_hAxk/видео.html .
a^b=a^c
a=0 ==>
Or a^p= 1, a0 : 2 variabkes ==> few casas :
p=0, a0
Or a=1 , p defined
Or a=-1 , p even
Solution:
x^x = 1/x | *x≠0 ⟹
x*x^x = 1 ⟹
x^(x+1) = 1^(1+1) ⟹
x=1 because the same operations are done with x and with 1.
I just wrote 1/x as x^-1 and took the log with base x on both sides so i got x=-1
Did you remember that I give you that problem, thanks a lot, but actualy it was
x^(x+1)=10
But thanks in any way!!!!
Loves and prayers from Ecuador!!!
Thank you! Likewise! 🥰
Too easy, thanks to log base x.
x^x = 1/x = x^(-1) => x = -1
1^(anything) = 1 => x = 1
Took me longer to type than to solve 🥱
What if the question is x^x=x ?
x^x=x
(x-1)*ln(x)=0
x-1=0
x=1
But the answer should be x=1 or -1
How to find the x=-1 ?
Let's solve your equation step by step.
x^x = 1/x
Step 1: Multiply both sides by x.
x^(x+1) = 1
Step 2: Take x+1th root.
x = 1
Answer:
x = 1
Wow this is an amazing problem ...
Easy to prove that x= -1 or 1 are the only real solutions ...
But are there other complex solutions ?
Sir (e to the power y)to the power(e to the power y) = e to the power (y*(e to the power y)) = e to the power -y therefore e to the power y = -1 implies x = -1
However there is an exception that 1 to the power 1 equals to 1／1
Your first method is false:
Lnx(x+1)=0 = ==>
lnx=0 ==> x=1 ok
or x+1=0 ==> x=-1 but ln(-1)(1-1) undefined.
Second method was published 4months ago!
x=-1 is valid solution
@@SyberMath Your method with ln is false:
(x+1)lnx =0 gives only 1 as solution
But ln(-1) not defined.
The good method is:
x^p=1 , p=x+1 is equivalent to:
{• p=0 and x0 : -1 is solution,
• or x=1 and p exists: 1 is solution,
• or x=-1 and p EVEN: -1 is solution bcse p=-1+1=0 is EVEN }
This RULE is in:
ruclips.net/video/aGBe9ObqlwA/видео.html
@@Frank-kx4hc I'm sorry, Sir Frank, but zero is neither odd nor even. Remember what zero represents - the lack of value. The number line is discontinuous at zero.
If you want to prove something: remember that odd plus odd is even? or, put odd times odd is odd. If you use zero, you will see you can't prove zero to either odd, or even.
You see, if you apply number theory you will see the point.
@@Roq-stone
without zigzag:
• (-1)^0 =1: so x=-1 is solution of your own equation x^(x+1)=1 ,
• any number (2k+1) is odd for all k in Z,
• any number (2k) is even for all k in Z.
Thus no need to speak about : x=0, 0 is neither odd nor even, number theory,...
I knew that although x = -1 might have worked, that was extraneous due to the fact of ln(x). Plus I figure to use properties of ln(1/x) being ln1-lnx. This is not a challenging problem at all.
x=1
x=-1 aswell
@@Chriib okay
@@Chriib no
@@kirillonf.m.4713 well, yes. Just check the video.
@@Chriib It's true, if we solve this equation in integer numbers. But this isn't said. Then we solve in real numbers (R), but x>0
oh. its holly war of roots of equation
f(x)^g(x)=f(x)^h(x)
It's fascinating how people (most of them certainly smarter than me) avoids understanding what Sybermath is doing here (maybe he wont even agree with me. But I say he's more right than he will admit). Once you have the algebraic expression giving x=-1 as an additional solution. Then that's that. It is obviously correct and -- if you insist -- easily verified. It has nothing to do with ln(-1) not being defined. That might be interesting in itself but irrelevant here. Always trust algebra over some applied mathematics exp or something. The simpler and clearer definitions the more powerful the use of them. Being vague to the point of reaching no results isn't that productive. Even mathematicians need to be a LITTLE productive. Even pure ones.
👍
x^x = x^(-1)
so x = -1 is clearly a possible solution
Very elegant
Log isnt applicable because by doing so you are ignoring negative values of x
Huh? In this application, both results were revealed. It threw out 1 and -1 as real solutions using the ln approach. Didn't you get that?
Good video, nice explanation!
Thank you!
x = ±1
reading this is see x^(x) = x^(-1) so x === -1 then consider domain
X^X=1/X X=-1 X=1
x=1 (at (1;1))
Quite interesting problem.
Start with multiplying x on both sides.
x^x * x = 1
Since we have x^x * x^1, we can add those two exponents together.
x^(x+1) = 1
The interesting part here is that if we take the x+1th root on both sides we get that x= x+1 root of 1. But because any nth root of 1 is just equal to 1 (for n not equal to zero which grants undefined) we get that x = 1, which indeed is a solution.
If we then take log_x on both left and right-hand sides we get that x+1 = log_x(1) (assuming that in this case x is not 0 or 1) we can say that x+1 = 0 leaving us with x=-1 which is another solution.
I know this way of solving it is very moot as there are a lot of false assumptions such that logx(1) = 0 even though it isn’t always true. But funnily, it does give the two solutions that we want so…
x = 1 or -1
For Y = X^X the graph in negtive domain has (-1, -1) point which is the only intercourse with Y = 1/X
So the graph should show the (-1, -1) green point along with the positive domain curve line.
The way Desmos works is before even solving any equation it takes a look at the domain even though negative integers are allowed it will not yield a continuous function so just by convention x^x is defined for all x>0 and that’s why you don’t see a green dot.
Intercourse?😳
Right. When you first graph y=xˆx the 2nd quadrant shows some dots which are some of the valid points
Very easy question ,x=-1
Very easy, but it's ±1 not -1
@@Andre-he5ly both are true
@@harrymcgilligan5807 yeah which is why you have to give both answers
No, not quite. Technically, the symbols x^x is an abbreviation for the more correct expression exp(x·ln(x)), which is only defined for x > 0. This is a really bad habit people online have, of using the symbol b^a to mean exp(a·ln(b)), when actually, b^a means something completely different than exp(a·ln(b)), and they have different domains (b^a is defined for all nonzero b and integer a, exp(a·ln(b)) is only defined for b > 0 but all a).
@@angelmendez-rivera351 actually both are defined and real if b is negative and a is an integer. If b < 0 then ln(b) = ln(-b) + i*pi and so exp(a*ln(b)) = exp(a*ln(-b)) * exp(a*i*pi) = +- (-b)^a. Sign depends on if a is even or odd respectively and the expression is real because -b > 0.
x^x =1/x
x*ln(x) = - ln(x)
(x+1)*ln(x) =0
ln(x) = 0
x = e^0 = 1
x+1 = 0
x = -1
The two solutions are -1 and 1
Isn't it immediately obvious?
write 1/x as x^(-1). so the LHS is the same form as the RHS. Then it can only be that x=-1.
That’s it? 🧐
@@SyberMath I'm a physicist. We don't usually worry about trivial solutions.🤣
xlnx=-lnx...x=-1...ma è valido?
how about lnx=0?
@@ngdluu ln(x)=0 - legal:
e^(ln(x))=e^0
x=e^0=1
but ln(-1) is not defined on real set.
if we use this treak, we can both part of every equation multiply by zero and than got 0=0, so cool, x € R. details not matter.
No it's not valid as it stands (assuming we are working in the real numbers). That's why SyberMath recommended using the absolute value. I suggest you see my comment for a solution using the absolute value.
nicee
x^x = 1/x
x^(x+1) = 1
(x+1) ln x = 0
Or x+1 = 0, or ln x = 0
x = -1 or x = 1
x^x=x^-1
x=1
х + 1 = log x, 1, so if x ≠ 1 or 0, log x, 1 = 0 and x = -1, else x = 1. 1¹ = 1/1 and -1^-1 = 1/-1
x=1,-1
-1 by inspection
Answer: x^2-1=0
This is the first time I’m going to disagree with you. Zero is neither odd nor even. Zero is not a member of the number line. Zero is a hole.
Your method with ln is false:
(x+1)lnx =0 gives only 1 as solution
But ln(-1) not defined.
The good method is:
x^p=1 , p=x+1 is equivalent to:
{• p=0 and x0 : -1 is solution,
• or x=1 and p exists: 1 is solution,
• or x=-1 and p EVEN: -1 is solution bcse p=-1+1=0 is EVEN }
This RULE is in:
ruclips.net/video/aGBe9ObqlwA/видео.html
With the absolute value, x=-1 is also valid
@@SyberMath
yes but the equation is (x+1)lnx=0, not (x+1)ln|x|=0.
x={-1;1}
I took a guess and said x = 1 and it worked.
x=1
x=1