X raised to the power of X equals Y. Given Y, how to calculate X?
- Thread starter dim35
- Start date
f(X) = Y = X^X
What is the domain of f(X) (given or implied) → consequently what is the domain of f(X)?
If Y=10, I can find X via trial and error to be 2.5061841456 (discounting rounding errors). I would like a formula to work it backwards.
The last line of response #2 should be corrected as shown above (↑).
Last edited:
The domain for both X & Y are positive numbers greater than zero. Sorry for the delay in replies but as a new member my posts need to be approved before you can see them.
For these types of functions, I d not know of any simple algebraic method to solve for 'x' (exactly) - as inverse of the given function. The only way I can suggest is to use graphical method or numerical method to approximate the solution.
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,087
This can't be done using elementary functions.
You can use the Lambert W function to solve [imath]y = x^x[/imath]. Take the natural log of both sides, then replace [imath]x[/imath] with [imath]x^{\ln(x)}[/imath].
In practice, though, you can just use any numerical approximation method, such as the midpoint method, to find x.
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,087
Not quite. Did you not read answer #8??
There are many methods of numerical approximation besides mere random trial and error. And there is a formula of sorts, if you have a way to evaluate W.
But if you describe all numerical methods as "trial and error", and want a formula using only elementary functions, then you are correct (but you are just restating, less clearly, what I already said).
Steven G
Elite Member
- Joined
- Dec 30, 2014
- Messages
- 14,383
Did you not read my earlier post (#3). It is easily seen that if y=4, then x=2 (no zeroing in!). If y=27, then x=3 immediately. I used no formula and did not zero in to get my answer. Now I never said that for all values of y you can get an immediate answer. You on the other hand said that the only way to solve this problem is by trial and error--that is not always true.
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,087
So your method in this case is not "trial and error", but "trial and success", better known as "solving by inspection".
That's useful for the very easiest problems ... or rather, it makes some problems very easy, when you just recognize a fact that you have seen before, just as some divisions are very easy, if you know your multiplication table.
But I was curious:
Did you mean to say, "you can"? If that's a typo, it certainly made it harder to follow your thinking.