Thou Shalt Find Roots
How many real roots does the equation
x
x
=
(
x
)
x
have?
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9 solutions
Why x=0 is not a solution? I know that 0^ 0 is indeterminate, but always 0^0=0^0.
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Not quite! Indeterminate forms are not equal, even if they look exactly the same; that is, if 0 0 = 0 0 , then it would imply 0 0 − 0 0 = 0 , which is false since 0 0 − 0 0 is indeterminate.
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Why is 0 0 − 0 0 indeterminate? 0 0 equals infinity, and logically, infinity minus infinity equals zero, doesn't it?
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@Jehad Aly – Infinity minus infinity doesn't equal 0. Because you could double the first infinity and get 2inf-inf=inf-inf=0, but 2inf-inf=inf+(inf-inf)=inf+0=inf, thus 0=inf. That's why inf-inf is also an indeterminate form.
I'd argue that 0 0 = 0 0 1 , then we cannot determine its result
Actually, 0 0 is undefined.
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Some people use different definitions to describe indeterminate and undefined, but the most common definition that I've seen at a university level is that indeterminate refers to a form that can have several reasonable values. 0 0 , for example, could sensibly be 0 or 1 based on the two most obvious approaches: 0 x = 0 , x ∈ R + , and x 0 = 1 , x ∈ R − { 0 } . By the definition above, it is then indeterminate.
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@Caleb Townsend – But zero doesn't have an inverse, so that first one should be 0 x = 0 , x ∈ R + .
And the reason it is undefined at the "university level" is because they use the definition a x : = e x l n a , so 0 0 would be e 0 l n 0 , which is undefined because zero doesn't have a natural log.
It's nothing to do with indeterminate forms, which only apply to limits.
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@Whitney Clark – If a x is defined as e x ln a , how come 0 x = 0 for positive x ? ;)
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@Otto Bretscher – That's a good question. All I know is, zero squared or cubed is zero, the square root and cube root of zero is zero, and so on.
@Whitney Clark – ln 0 gives negative infinity, and multiplying an infinity of any sign with a 0 gives the indeterminate form in limits.
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@Paul Paul – I wasn't talking about limits, just plain exponentiation.
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@Whitney Clark – Whenever I say the result is infinite, then restricting myself to normal reals obviously means that it becomes meaningless.
@Caleb Townsend – There are limits where 0 0 could get anything.
In algebra 0 0 means 1 . So in this point of view 0 is a solution. It is a matter of convention we use. For istance N is considered as the set of all positive (strictly) integers by someone, and the set of all non negative integers by other. Wheter 0 belongs to N or not depends on our assumption...
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In any commutative unit ring ( R , + , ⋅ ) anyone can define by recursion, for every a ∈ R
a 0 = e m a n + 1 = a ⋅ a n for every poitive integer n
where e m is the neutral element for the operation ⋅ .
It is right that 1 is an obvious root.. But is there a mathematical way to get it?
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You can do it this way, Jehad: Take ln on both sides to find x ln x = x ln ( x ) = 2 x ln x , for x > 0 . Multiplying with x 2 , we find ( x − 2 ) ( ln x ) = 0 . The two solutions are x = 4 and x = 1 .
gud and simple
Taking natural logarithm from both sides yields:
x ln x = x ln x
⇔ ln x = 2 x ln x
⇔ ( x − 2 ) ln x = 0
⇔ ln x = 0 ∨ x = 2
⇔ x = 1 ∨ x = 4
Hence, there are two real roots.
Why don't you use log? is that okay to use log? what is the difference with use ln?
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That's the way I used to be taught - they're identical. I know it's only a naming convention, but somehow I find the notation ln more satisfying to the eye. Just a personal taste you know.
whenever i confuse i take log
x^(x^1/2) = x^(x/2)
x = 1 ............................................... (1)
or
x^1/2 = x/2
4x = x^2
x = 0 .................... refused
x = 4 .............................................(2)
This equation has only two real roots , 1 , 4
I found it easier to let x = y^2. We then get (y^2)^(sqareroot(y^2)) =( y^2)^y = y^2y on the left. On the right we get (squareroot(y^2))^(y^2) = y^(y^2). Thus, y^2y = y^(y^2), leaving 2y=y^2. When is this true? When y^2-2y = 0, or y(y-2)=0. Solving, y=0 or y=2, so x=y^2=0 or 4.
take log(base x) on both sides....you get: sqrt(x)=x/2 => x^2-4x=0 => x=0 or x=4
x=0 is invalid and you missed x=1 by taking log base x.
As,x to the power rootx = rootx to the power x; so, x to the power rootx = x to the power x/2; As the bases are same; rootx= x/2; Squaring both sides, x=x^2/4; x^2=4x; x^2-4x=0; x=0, x=4;
x^sqrt (x)={sqrt (x)}^x, as given... Raising both the sides by the power of sqrt (x)^(-1), we have, x={sqrt (x)}^sqrt (x). we get that only the values for x=1 & x=4 satisfy the equation.
Take log on either side and reduce the expression. (x^(1/2)) log x = x log(x^(1/2)) (x^(1/2)) log x = x (1/2) log x logx=0 or (x^(1/2)) = x (1/2) i.e. x=1 or x=4 Therefore 2 roots.
x = 1 is an obvious root.
The given equation can be re-written as x x = x 2 1 x . Equating the exponents we have x = 2 1 x or x ( x − 4 ) = 0 .
But 0 isn't a root of the original equation. Hence there are only two roots: 1 & 4