Simply Exponential
x x = x Find the number of distinct real solutions of the equation above.
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12 solutions
Moderator note:
Great case checking of x y = 1 . It's a pretty common olympiad problem, and using this approach is often helpful in those cases.
You guys need to get your Facebook presence sorted out. The question in Facebook is often not the same as in here, causing wrong answers. The Facebook question was "Can you solve for X?" while here you ask for the amount of solutions :)
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That's right. I got it wrong because I only looked at the question on FB and simply glanced at the question here.
On FB it says "Can you solve for X?" Which is not the same as finding the number of real distinct solutions.
Apparently the Brilliant.org person posting these questions to FB doesn't understand the question.
Yes. This.
I made that same mistake.
Yep I felt stupid when I got it wrong
That's very nice!
what about 1^1=1?
Si aplico logartmo en base x (sea Logx) a ambos lados de la igualda obtengo: Logx X^X = Logx X XLogx X = Logx X Pero; Logx X = 1 entonces X =1 Esta es la única respuesta o valor de X posible .. this is my answer.
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si x^x<=0 no puedes aplicar el logaritmo
isn't 1¹=1???
the problem states x^x and not x^(x-1) the solution is 1
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Yes, but you have to rearrange the equation first. You can't deal with x on both sides of the equation if you want to work out all possible solutions, which is what we're asked to do. Have a look at Ivan Koswara's comment more closely.
If I take log of both the sides of the equation, I get xlogx = logx, or, (x-1)logx = 0. Hence, x = 1, or, logx = 0, giving, x = 1, again! In this method, I don't get the other root of the equation, viz., x = -1. So, what is wrong? Or, where am I wrong?
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When you took the log of both sides you applied a condition that x>0. This is because for real solutions x has to be positive.
Did it the same way, forgot the negative part :)
But, if the exponent is an even integer, it's different; the problem has all 3 positions equal at simply 'x'. I don't see how you can plug in any even integer, which by def is not -1 or 1, into the exponent 'x'. Please explain further; I'm probably missing the point of We check that the exponent is -2, which is an even integer.
The answer CANNOT be -1 bcoz base can never be negative. Hence, there is only one answer and it is 1.
it tells me the awnser is 2 and im like no its 1 because 2 to the power 2 is 8 so its clearly 1 cause the only number i can think off that would fit that variable timesed by itself and equals itself is 1 cause 1x1 is 1
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look at the question: how many real solutions are there? and not: what is the solution?
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I see what you mean - I didn't read the question correctly. So there are two solutions: 1 and -1. What about 0? Does it not count because it is not "real?"
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@Jason Carter – Am I understanding the question correctly? 1 and -1?
@Jason Carter – 0 wouldn't be correct, as 0^0=1, as any number^0 is 1. So only 2 solutions
He didn't misunderstand the question. The FB link to this question says "Can you solve for X?" Like me, he probably thought "What an easy question!" and didn't look at the question a second time once he arrived here. Obviously this is different, but usually the FB post actually has the proper question on it.
-1 is a second a solution, but that wasn't on the list of answers when you're expecting a simple "solve for X" question. From the FB posted Question to here, one would simply be looking for positive integers. He was misled.
The problem is x^x=x hence the solution is 1. The explanation states x^(x-1), which is a different problem altigether
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@Carlos da Cunha – The problem states x^x=x, and the explanation transforms that equation to x^(x-1)=1. There is no conflict.
2 to the power of 2 is 4
No, it tells you there are 2 solutions, since the question is "How many solutions there are" and not the value of X
I agree. I think number 1 fits. 2^2 = 8
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- _ - 2 to the second power is 4....
Now I got it. I just misunderstood the question.
x x = x
ln x x = ln x
x ln x = ln x
ln x x − 1 = 0
x − 1 = 0 → x = 1
Note that each time that we solve this kind of equation, we should consider positive and negative answers. Answer − 1 , 1
Dividing both sides by x, x x x = 1 = x 0 And x x − 1 = x 0 ⇒ x − 1 = 0
Should we really need logarithms?
After reading the whole page I came to this conclusion, "Your's was the best solution!"
I think you can also prove it this way-
start substituting the values for the equation,
let x be 0. therefore 0^0 is not equal to 0.
let x be 1. therefore 1^1 is equal to 1.
let x be -1. therefore -1^-1 is equal to -1.
let x be 2. therefore 2^2 is not equal to 2.
let x be -2. therefore -2^-2 is not equal to -2.
similarly it is not possible for 3, -3, 4, -4, 5,........................... therefore it is 1 and -1 only!
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That is not a proof. Plugging in random values only proves it for those values, not in general. It just so happens that this will continue to work, but you have failed to prove it in general.
How can 0x0 not equal 0? EVERYTHING times 0 ALWAYS equals zero therefore 0x0 is a valid answer.
Also, when multiplying negative numbers, the result is ALWAYS positive therefore -1 x -1 equals +1, NOT -1. -1 does NOT satisfy the equation.
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It does, you are wrong. 0^0 is an indeterminate form. x^0 = 1, but 0^x = 0, so there is no way to determine what 0^0 will be equal to, making it indeterminate. This is a subject of study in calculus when doing L'Hopital's Rule of limits. x = -1 is, however, a solution:
-1^-1 = 1/-1 = -1.
Any number brought to the power of -1 results in the reciprocal of that number. 1 over -1 is indeed -1.
0*0 is not equal to 0^0. Anything to the power of zero equals 1 and anything with zero as the base has zero as the answer. So the answer can be either, but we don't which, so it is indeterminate.
-1 x -1 is the same as -1^2, not -1^-1. Anything to the power of a negative basically means the reciprocal of the base to the power. E.g. x^-1 = 1/ (x^1) . 6^-2 = 1/ (6^2) = 1/12
Just to help out. He didn't mean 0x0. He meant 0^0 or 0 to the power of zero
Yeah, only this is not a multiplication, is it?
I agree strongly with this, that -1 can not be the solution of this equation. The same has been demonstrated using method of IN. But the author has forced -1 to be the solution. -1 n.e -1 its 1=1 the minus sign always cancel.
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@Vikram Narang – Well it is not IN, but it is l n which is the natural logarithm.
@Vikram Narang – You don't think 1/-1 = -1 ?
My aproach was graphing two function y=x and z=x^x and find their intersection.
by indices x^x = x^1 the x cancels the other x leaving the equation as x=1
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that what i thought as well how was the answer 2?
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They were looking for "the number of distinct real solutions", not the solution to the problem. I made the same mistake. Also, the answers are 1 and -1, thus 2 solutions.
Nice solution, by the way: its l n not I n
Paola even I agree with the two solution , 1 and -1 ,I consider your proof defective, since you are dividing by 0 or by ln(-1)
We can only consider the negative of it if it is odd but not in general.
Ln(-1)=DNE
You can not use ln on your equations because by doing that, you exclude all the negative possibilities for the answer, since ln of a negative number does not exist. When you state ln(x), you are assuming x>0
If x = − 1 then what is value of the secont equation?
ln ( − 1 − 1 ) = ln ( − 1 ) which is not defined.
The Ln of negative numbers is not defined. So -1 has been forced as the solution. Rather this way we can demonstrate that -1 cannot be the solution.
Each time you take logarithm you go on with the assumption that the number is not negative and also not zero. Now to say that after solving via log method, if you get 1 then -1 would also be the answer is wrong at many levels.. :P
The last step is wrong. lnx^x-1=0 or x^x-1=0 leads to 2 solutions, not 1. x=-1 is not explained, like it is in another post, in which solution is correct and more complete.
Anyway, I am confused about something. Why can't we divide both sides of xlnx=lnx with lnx? That would lead to x=1 only. What is wrong with this division? Why does it produce only one solution?
EDIT: nvm, Prasun Biswas explains why perfectly
x should not be negative because log of negative number is not possible then why x=-1.
But Paola if we put x =-1 then lnx will be not defined as in logarithm given log a to the base b a>0 and b>0 and not equal to 1
I'm with you. The answer says is it 2, which is plainly incorrect. Either 1 or -1 is a valid solution, and given only 1 is offered, I chose 1. It said 2. Absolutely and evidently wrong. +1
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The question is asking to find how many solutions there are. There are 2 solutions, being 1 and -1. The answer of 2 is correct.
x = 1 is invalid solution for the equation x ln x = ln x ; you cannot multiply or divide by 0. answer is 1 and value of x is infinity
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1^1 = 1 , so one is a valid solution to x^x = x, and it is also a valid solution to x ln x = ln x, since 1 * ln 1 = 1 * 0 = 0 = ln 1.
x x = x
Computing the x-th root x = x x
Thus, x x = x x .
Taking the logarithm, x lo g ( x ) = x 1 lo g ( x )
Or, ( x − x 1 ) lo g ( x ) = 0
Or, ( x 2 − 1 ) lo g ( x ) = 0 (as x = 0 )
Which gives the solution as x = − 1 , + 1
The x-th root has no meaning if x is negative Although x = -1 is a solution for the given equation, but x = -1 is a refused solution using your own method
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The x t h root if x is negative say x = − y where y > 0 is simply the inverse of the y t h root.
x=x^x ln(x)=ln(x^x) ln(x)=x•ln(x) x=ln(x)/ln(x) x=1
Proud of you sir
I have a doubt in your procedure.
What is the domain of definition of the following equation?
( x − x 1 ) lo g ( x ) = 0 . . . ( 1 . 1 )
I think it is x > 0 . What do you think?
If I am correct then, by solving equation ( 1 . 1 ) for x we cannot conclude that x = − 1 . Or can we?
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The solution is either x − x 1 = 0 or lo g ( x ) = 0 . The log function is also defined for negative numbers, though the logarithmic values would be imaginary numbers. lo g ( − 1 ) = i π .
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But in the question we have for "real" numbers. Then still can we say that ( x − x 1 ) lo g ( x ) = 0 for x = − 1 ?
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@Soumo Mukherjee – will you please find out error xlnx=lnx lnx(x-1)=o if lnx=o=> x=1 if x-1=0 => x=1
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@Shivam Verma – I wanna know this too. ln is also defined for negative numbers, but with imaginery values. So you would also have to search negative values of x. So, since -1 should be a solution, apparently (-2) ln(-1) =- 6.28318531 i=0
EDIT: Ok, Prasun Biswas explains this perfectly. ln(x^x)=/=xlnx, for negative x.
What about x = 0?
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@Shashank Rammoorthy – If x=o then 0^0=1 Not equals 0
@Shashank Rammoorthy – Because 0^0=1
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@Kallol Dhar – 0 x = 0 for all non-negative (x)
y 0 = 1 for all y = 0
There is therefore an ambiguity when trying to define 0 0 .
Ok, thanks for clarifying my doubt. Your solution deserves an upvote then. :)
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@Soumo Mukherjee – For ( x − x 1 ) lo g ( x ) = 0 , yes...the domain is x > 0 .
The solution should have continued this way:
x x = x x
x x x x = 1
x ( x − x 1 ) = 1
Thus, x − x 1 = 0
x 2 = 1
x = ± 1
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@Roman Frago – I think it's reasonable that (-1)^-1 is -1, but I wonder if the function y=x^x is defined for non-positive x. if so, can you show me the graph of this function for negative x ???? !!!!!
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@Mohamed El Tohfa – You can't 'see' the graph when x<0 Because it is imaginary.
The domain is x = 0 .
I marked 1, but told e I was wrong and the answer is 2. is that possible?
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Because we all didn't read thoroughly the question, it asked for the number of answers as in the amount of answers in Real Numbers, not for the answer.
simpiy first power of 1 is 1 so x=1 what is wrong with it
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Nothing wrong but you over looked another possible solution which is by rearranging the equation to be:
x ( x x − 1 − 1 ) = 0
{ x = 0 x x − 1 − 1 = 0
be aware x = 0 makes the function undefined
you'll find out that
x x − 1 − 1 = 0
x x − 1 = 1
has 2 solutions "1" which is the obvious one
and "-1" as ( − 1 ) − 2 = − 1 2 1 = 1
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so how comes the answer 2????
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@Halbert Joshi – Because he asks about the number of solution!! and you have 2 solutions 1 and -1
Indeed....!
This is not. The correct solution
Damn it I hate my stupidity
After figuring out that -1 also can work as a solution and even proving it to myself by rearranging the equation to be:
x ( x x − 1 − 1 ) = 0
{ x = 0 x x − 1 − 1 = 0
I just got lost in the fact that x=0 makes the function undefined and did some quick researches about it to make sure that I'm not mixing something, and when I came back I just clicked 1 !!
I hate it when that happen
Or by simply dividing which gives x 2 = 1 , thus x = 1 , − 1
nice solution:-)
@Prasun Biswas @Soumo Mukherjee Would it be possible for you to kindly tell me my mistake? x x = x ⇒ l n ( x x ) = l n ( x ) x ln ( x ) = ln ( x ) ⟹ x = 1 Many thanks:)
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A naive argument is that taking logarithm is only defined for positive reals, so when you take logarithm, you're restricting the domain of the functions on both sides to positive real domain. Because of this, the case of x = ( − 1 ) which is a solution gets overlooked.
Now, there comes a time when you learn that logarithm is actually defined for all non-zero complex values . So, we improve our argument to make it rigorous.
In complex logarithm, the identity ln ( x y ) = y ln ( x ) that you used in your second line doesn't necessarily hold true, which is the main reason for the error. Here's the demonstration:
ln ( − 1 ) = i π and ln ( ( − 1 ) − 1 ) = ln ( − 1 ) = i π = ( − i π )
The complex logarithm is actually the inverse function of the complex exponential function (Euler's formula).
Moral of the story: Try to use logarithms carefully. In this question, there is no need to use logarithms. A simple usage of zero product property does the trick.
x x = x ⟺ x x − x = 0 ⟹ x ( x x − 1 − 1 ) = 0 ⟹ { x = 0 x x − 1 = 1 ⟹ x = 1 , ( − 1 )
Now, note that x = 0 makes the original equation undefined and we can easily see that the other two solutions are valid. Hence the answer follows.
The case of x x − 1 = 1 is actually dealt with a number-theoretic fact which is:
If a b = 1 , we must have one of the following three cases:
(i) a = 1 , (ii) b = 0 , a = 0 or (iii) a = ( − 1 ) , b = 2 k for some integer k .
you don't need to use that logarithm because you take it wrong from the beginning. \sqrt[x]{x} is not equal to x. in simplicity, just use your logic. is 2^2 equal to 2? of course it's not. the answer is not two, use ur logic here, sir.
I do not fully agree with your reasoning by taking the x-th root you are changing the range of the domain of the problem as enunciated, for instance 2^4=16. Now we take the square root of both sides we get 2^2=+-4. Actually you are mirroring the functions z= x^x and y=x over OY axis Besides 1 and -1 each one are doble roots since y=x is tangent to x^x at x=1, so rigorously there are two pair of double roots
log(-1)is not defined
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log(-1) is not defined in a base of numbers greater than zero. You might have thought this base is "e". Actually the base of this log is x. Depending of the value for x, log(-1) could yet exist, if x<0, which is the case.
The answer is right, but probably the explanation given is not right. The problem is that we are taking log to get to the answer. Hence we are ignoring the negative solutions for x. Good question.
This makes no sense, i'm taking algebra and as far as i can tell, if x to the x power equals x then how is 1 wrong because 1 to the power of 1 equals 1 it could be an answer
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It's because it's asking how many solutions there are. The answer is 2 because it could be 1 or -1.
@Janardhanan Sivaramakrishnan Sir, the bases in exponential functions cannot be negative or zero , they have to be positive , and thus I think -1 has to be omitted. Pls correct me if I am wrong
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The base of an exponential maybe be negative. For example, (-2)^2 is 4, and (-3)^3 is -27. The domain of an exponential function is all real numbers. You may be confusing this for logarithmic functions, of which the domain is all positive real numbers, assuming the desired range is real. That said, if the allowed range may be imaginary, the domain of logarithmic functions is simply non-zero.
How this i can solve it with powers so , if base =base so power =power so x=1 one solution
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but you over looked another possible solution which is by rearranging the equation to be:
x ( x x − 1 − 1 ) = 0
{ x = 0 x x − 1 − 1 = 0
be aware x = 0 makes the function undefined
you'll find out that
x x − 1 − 1 = 0
x x − 1 = 1
has 2 solutions "1" which is the obvious one
and "-1" as ( − 1 ) − 2 = − 1 2 1 = 1
If i use base =base so power =power so x=1 one solution
yes good solutionm. but u better realise that log negative is illegal
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Taking the logarithm of a negative number is not a crime. It is not illegal. It is just not a real number.
Is this correct ?
from the question x x = x
but x x is not equal to x x
for example 2 2 is not equal to 2 2
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Simple answer : x = 2 .
However, 1 1 = 1 and − 1 − 1 = − 1
just use the simplest method; try all the answers. 3^3=27, 2^2=4, 0^0=1, 1^1=1. thus the only option that meet the condition is 1 (except if there are other properties in that equation that i haven't learned) (also sorry if the english is bad)
I put in the answer of 1 and it said I was wrong and that it was two.... I think you need to fix something
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The question asks for the number of real solutions. It does not ask for the solutions themselves.
I dont know where to comment because I am here with my wrong answer 1, but if the answer 2 is true then it is shaking the basis of fundmentals of mathematics.... 2^2=2......wow
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It is number of solution and not a root of the eqn if u read the question properly.
it can't be -1 because log(-1) is not defined.
no it's right... sorry it's a solution of the given equation.
How can x be - 1 when in your own proof you have used it as an argument for log function?
Can you explain the third line
The x-th root has no meaning if x is negative So x = -1 is a refused solution if we use your method, although x = -1 is a solution for the given equation
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Stop fighting for your x, there's plenty of girls out there. Hhaahah PEACE!
Not entirely true. The x-th root of x is undefined in the realm of real numbers for all EVEN values of x when x<0. It is certainly defined in the case of ODD values of x, even when x<0. This latter case contains the other solution, x= -1, making the total number of solutions 2, per the answer.
Graphically solving there is only one solution. Check https://graphsketch.com/?eqn1 color=1&eqn1 eqn=x%5Ex&eqn2 color=2&eqn2 eqn=x&eqn3 color=3&eqn3 eqn=&eqn4 color=4&eqn4 eqn=&eqn5 color=5&eqn5 eqn=&eqn6 color=6&eqn6 eqn=&x min=-5&x max=5&y min=-10.5&y max=10.5&x tick=1&y tick=1&x label freq=5&y label freq=5&do grid=0&do grid=1&bold labeled lines=0&bold labeled lines=1&line width=4&image w=850&image_h=525
But the answer is 2
You are wrong, and all this answers are wrong, the answer is 2, why BECAUSE IT ASKS FOR THE NUMBER OF REAL SOLUTIONS. not the solutions of the ecuation at all.
Also as a side note (not that they ask for) the solutions are not 1 and -1, they are 1 and 0 why? allow me to explain as follows:
You have x^x=x which is the same as x^x-x=0
You factorize one x and you get: x(x^(x-1) -1)=0
We will call the exponential x-1 y so you will have x(x^y -1)=0
Now you have to meet two conditions to have a result of 0 one is the x being 0 as it is alone and the othet one being x^y-1=0 and for that matter x^y=1 so it doesnt matter what power you use as 1 powered by anything is still one and the same goes for the rooting of 1.
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yes u are right, answer is 2. there are 2 nos. of real solution
x x = x x x − 1 = x 0 L e t ′ s j u s t m u l t i p l y t h e e x p o n e n t b y ( x + 1 ) x ( x − 1 ) ∗ ( x + 1 ) = x 0 ∗ ( x + 1 ) x 2 − 1 = 0 x = ( − 1 , + 1 )
I think we cannot introduce a power like this, otherwise we can do that for any power. Say (x-5). Then x^(x-1)=x^0 becomes x^(x-1)(x-5)=x^0(x-5). Comparing power, (x-1)(x-5)=0 which leads to x=1 or 5. Obviously, x=5 is not a solution to the original eqn x^x =x.
You cannot multiply variable to 0. For instance, if I multiply the exponent by (x+2) instead of (x+1), the answers will be 1 and -2, which is totally wrong.
1 and -1 are those two solutions.
1^1=1 and (-1)^(-1)=(-1). So, there are two solutions.
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I added the condition that x is real, so that we do not need to deal with complex exponentiation.
Right. Answer is 2 solutions: 1 and -1, not 0.
OK It may sound a bit mechanical, but why can't we do this?
x x = x ⇒ x lo g x = lo g x ⇒ lo g x ( x − 1 ) = 0
I haven't given much thought to it. But, I wanna figure out why are we losing solution that way?
A little help will be appreciated. ;)
Update!
Don't mind, got the glitch. By taking log we are reducing the domain of definition of the variable. So, after taking log both sides and not forgetting the lost domain, if we proceed and bring into play the left domain afterwards, we may find the full solution. Still want to know if we can get a standard method.
Ok so while the hot-pursuit I also got this:
x x x x . . . = x
Will this too have same solutions?
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x = 1 is definitely a solution, and x = − 1 appears to be a solution, (since the infinite power tower on the left-hand side would still be of the form ( − 1 ) − 1 = − 1 ), but I'm uncertain as to the convergence of an infinite power tower that involves negative values. I need to do some more research on this; WolframAlpha only gives x = 1 as a solution, but it's not an infallible resource.
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XD * infinite tower * hhaha. how did... lol. u r creative in naming things. :D
brian charlesworth working on my randomly thought question
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@Soumo Mukherjee – Haha. Well, "power tower", whether finite or infinite, is actually the standard terminology; apparently mathematicians do have a sense of humour. :)
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@Brian Charlesworth – Oh.. is that so? I didn't know that is called ''power tower'' and that it is a standard term. Well, now I do.
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@Soumo Mukherjee – Within the field of real numbers your infinite power tower, (or tetration), converges only for e − e ≤ x ≤ e e 1 , so x = − 1 is not in fact a valid solution. (WolframAlpha was right after all.) Things get a bit more interesting when we extend to the field of complex numbers. You can read more on this topic here .
Your "randomly thought question" was a good one. Thanks for asking it. :)
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@Brian Charlesworth – (^_^) that's motivating, thanks. But the credit goes to you. You were then one to spot something in the question, whereas I just gave it the value of a random thought.
if people are given at first to attempt the x to the power x problem, and then the problem concerning infinite power tower they will be more vulnerable to make mistakes. I will try it on some of my friends.
One more thing. How can I reach you if I have some doubt, or want to discuss some problem with you. Not every time I get a reply to my doubt-notes. I have thought that in my doubt-notes I would mention your name @brian charlesworth , so that you can get notified and if have some time then work on it and help me out. Just wanted to tell you beforehand.
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@Soumo Mukherjee – Mentioning my name in your "doubt-notes" is fine. I look forward to working on whatever questions you might have, (time permitted). Thanks for letting me know ahead of time. :)
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@Brian Charlesworth – I have been trying to mention you Here but it seems not to work. This is a problem that I posted recently. And my problem has been reported. I have replied to respective disputes. But I have not got any reply in turn. I want you to have a look at it. And tell me what you think. It is going to be about 24 hrs. And I fear my problem will be deleted which I think is not wrong on the whole.
If you get any time, plz look into it.
thanks
Update
Sorry to bother you . My mentioned trouble has been dealt with. And I am completely satisfied. :)
I am updating it because I think you might get notified that I have replied to you. And deleting my comment would be not a good idea. In short: plz ignore this comment, don't wanna waste more of your time.
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@Soumo Mukherjee – I've had a look at your problem and I still think that there might be an issue with the wording. Both solutions to x + x 1 = 1 are complex, so the condition that 0 < x has no meaning. We can solve for x 2 + x 2 1 given x + x 1 = 1 , but if we also require that 0 < x then technically there is no solution. So as presently worded the answer would be "none of these", but without the 0 < x condition the answer would be − 1 . I haven't answered the question yet because I'm not sure if you meant this to be a "trick" question or not, which is one of the main reasons I always hesitate answering multiple-choice questions.
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@Brian Charlesworth – yup, the question is 'a trick question'. But I got my disputes resolved. It was flagged and had more than 3 upvotes in a dispute. I replied to them but still they were there. I was worried I would fail to express what I wanted and they would delete it. Only after giving enough thought did I put that question. Before doing that I wanted to ask you that, is i>0?, where i square is equal to -1. I have asked this to someone else in Brilliant, but I got an answer but not a satisfactory answer. By square root if we mean a the principal square root which is positive why not i>0.
But I realized later that I need not to know whether i>0 to post it and can post the question anyway.
Thanks a lot for having a look into it, even I told you, afterwards, not to. The moment that I commented here just after 4 to 5 minutes later they replied to me.
Thanks :)
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@Soumo Mukherjee – Whereas real numbers can be placed on a line, complex numbers exist in the complex plane, such that the coordinates ( x , y ) corresponds to the complex number z = x + i y .
Now say we're back in the real plane. It wouldn't make sense to say that the point ( − 1 , − 3 ) "is greater than" the point ( 1 , 1 ) , but it would make sense to say that the magnitude of the former is greater than that of the latter.
The same goes for complex numbers. It wouldn't mean anything to say that z 1 = a + i b is greater than z 2 = c + i d , but we could say that ∣ z 1 ∣ > ∣ z 2 ∣ if and only if a 2 + b 2 > c 2 + d 2 .
So i = ( 0 + i ∗ 1 ) > 0 doesn't mean anything, but ∣ i ∣ = 0 2 + 1 2 = 1 > 0 is a valid statement.
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@Brian Charlesworth – Let the last para be sentenced to death (-_-)
@Brian Charlesworth – The reason this converges properly is that even roots of negative numbers are not real. However, odd roots of negative numbers are perfectly defined. Because of this, Wolfram Alpha's algorithm likely missed the potential for odd negative values. Taking this into consideration allows for the inclusion of -1 as a potential solution, which indeed it is.
yes,sir...!!
You already pointed it out in your first comment, it limits the domain to x > 0 .
how 0 ? isn't 0^0=1.
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No, 0^0 is undefined.
In either case, 0 is not a solution to this equation.
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If we apply limit to the expression x tends to zero it is coming out to be zero! Then would we count it as one solution or not?
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@Nimesh Patodi – From which direction?
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@Roman Frago – Both sides
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@Nimesh Patodi – x x approaches 1 not 0 when x approaches 0 from the positive side.
Also x x becomes imaginary when x approaches 0 from the negative side. x x is imaginary when − 1 < x < 0 .
I don't understand how it can be 2 ??
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@Daniel Abdou – Without any mathematical proof, it is obvious that it can only be 1 or -1.
What are you talking about "undefined"? ANYTHING x 0 ALWAYS equals 0 therefore 0 satisfies the equation.
How the hell can 0 squared equal 1? Anything times 0 =0
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0^0 does not equal 0x0, just as 3^3 doesn't equal 3x3.
How ridiculous i am, i thought it was x . x instead of x x :(
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Even for x 2 = x you will get 2 solution.
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Do you mean x . x = x ? I think it has 3 solutions? {-1, 0, 1} ? That's my mistake
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@Andronikus Lumembang – Nope. Minus x on both sides, and you get x ( x − 1 ) = 0 after factorisation which means x = 0 , 1 .
@Andronikus Lumembang – (-1)*(-1) is not (-1)
x^x = x now, taking log both sides,we get
xlog(x) = log(x) or x=1 and never forget to include its additive inverse i.e. -1 in it in this way we get two real solutions of the equation
If x is greater than 0. Then x=1 and if x is less than 0then x=-1 So total root are +-1ie 2
-1 and +1 are the only two possible values.
powers of negatives still produce the same result in negative form therefore 1 to the power of 1 =1 and negative 1 also is negative 1, 2 possible answers
We have two cases 1º)x>0 we can will conclude that x=1 2º)x<0 So x=-y,and y>0 (-y)^(-y)=-y (-1)^(-y-1)=y^(y+1) Since y is a real number,so (-1)^(-y-1) must be So,(-1)^(-y-1) must be 1 and -y-1 must be 2k,where k is a 0,1,2,3,... Then we have y^(y+1)=1 y>0 Which give y=1 So x=-1 And sorry for my bad english
First, observe that x = 0 is not a solution, as 0 0 = 0 . (There are two most common values for 0 0 , being either 1 or undefined, and neither is equal to 0 .) Thus we can divide both sides by x , giving x x x = x x − 1 on the left hand side, and x x = 1 on the right hand side. Thus x x − 1 = 1 .
There are three ways for a real number raised to the power of a real number to be 1 :
This gives x = 1 , which is a solution (since x x − 1 = 1 1 − 1 = 1 0 = 1 .
This gives x − 1 = 0 , or x = 1 again.
This gives x = − 1 . We check that the exponent is x − 1 = − 1 − 1 = − 2 , which is an even integer, so x x − 1 = ( − 1 ) − 2 = 1 , and so x = − 1 is a solution.
Thus there are 2 real solutions: 1 , − 1 .