Solve explicitly for x: a^(a-x) = x . Betcha cant.?

Question

Use whatever functions you want. Just solve explicitly for x.

ag_iitkgp You didnt solve for x

fatal_flaw_d...

that may be correct, you're on the right track but LambertW of x is not explicit in x, it is implicit

prbsparx...
the equation as written is an IMPLICIT

prbsparx
as i was saying, as written it is an implicit function.
it CAN be written as x=f(a)

TO :themuffinking01

x is a function of a ; it is not a constant.
as an example,
a=2
x=1.45699955913459
works.

Answer 1

Here's what I know so far:

If f(a) is the explicit function then,

* for a < 0, f(a) is not continuous.
* for 0 <= a < b, is not in the domain of the function .
where 'b' is a value such that [ln(b)*b^b = -1/e] [eq. 1](b is about .6079). I've not been able to reduce this equation any further to get the real value. I'll show later how I got to this.
* for b <= a < 1, the function curls around on itself like a hyperbolic function with the top portion of the x values having an asymptote that approches 1 from the left. In other words, over this range there are 2 values for x that satisfy the equation for a single value for 'a'
* for a > 1, the function has the appearance of an exponential function, but always remains below the line a=x. This is also clear by inspection - if x were ever greater than a then a-x would be negative. Since a >= 1, a^(a-x) would have to be less than 1. But x>a>=1. So x can never be greater than a. Also, for x=a, you get a^0 which is 1. The only value for a that satisfies this 1. So, for a > 1, x must always be less than a. All of this implies that f(a) has the form u^v, where u and v are functions of a and lim(v) as a->inf is a constant in the range (0,1]. From the graph it looks like 1. I've determined through trial and error that a close approximation (at least for a < 15) is a^(1-1/a^1.25).

Partial proof:

Re-write the equation as f(x) = a^(a-x) - x [eq. 2]

f(x), written this way is a parabola for 0 < a < 1. For a, such that 0 < a < b, the low point in the parabola is a value for x such that f(x) > 0 - meaining the f(x) doesn't cross the x axis. This of course means there are no values for x that satisfy a^(a-x)=x over this range. For b < a < 1, f(x) is a parabola that crosses the x axis at two points - meaning there are two values for x that satisfy a^(a-x)=x for a given value for 'a'.
Example:
a=.7, x can be either be ~1.1916 or ~5.4581.

This can also be shown as:

d f(x)
------- = -a^(a-x)*ln(a) - 1. [eq. 3]
dx

Setting d/dx = 0, we get:
x = a - ln(-1/ln(a)) / ln(a) [eq. 4]
which is only valid and positive for 0 < a < 1.
which means f(x) has a minimum at this point (for 0 < a < 1).

To get 'b', I just need to find the value for 'a' < 1 such that f(x) = 0 at the point where df/dx = 0. To do that I:
Used [eq. 4] for x in [eq. 2] and reduced - ending up with [eq. 1]. I'm not going to show the work here - it was harry enough looking using pen and paper.

Answer 2

Sure. X=1.

Because the only time this works is X = 1, A = 1.

Here, look:

suppose x>a>1.
a-x < 0 so a^(a-x) < a, and because a^(a-x)
suppose x a-x > 0, and so a^(a-x) > a, and because a^(a-x)>a>x, a^(a-x) cannot be x and so the function does not work.

Suppose x = a != 0.
a-x = 0, so a^(a-x) = 1, and this means x must be 1. [this is the case that works].

I don't really know about the area between 0 and 1, but I'm pretty sure it doesn't work there either.

Answer 3

are we allowed to use logs? or calculators?

This seems to be a trick question, I supect the answer is simply a^(a-x)

you may be able to get all the x's on one side but you can't solve it to be an exact number... not without two equations and at least one of those having a numerical digit in it

Answer 4

My answer is X = 0

Answer 5

x = LambertW(ln(a) * a^a) / ln(a)
Where the LambertW function satisfies
LambertW(x) * exp(LambertW(x)) = x


One other thing we can note about this function is that:
as a -> infinity, x -> a-1 from above.

Answer 6

attempt the definition of logarithm. That huge type x such that once you improve 5 to it, you get one hundred, is defined to be: log(5) one hundred, the bottom-5 logarithm of one hundred. it is your answer, log(5) one hundred.

Answer 7

Let x = a^y

So, a-a^y = y

Let y= log z

So, a-z = log z

or z+logz = a

Answer 8

Betcha I don't wanna.