0
$\begingroup$

So basically I’m trying find the inverse to this formula and I’m having trouble getting it could anyone help me out I would appreciate it thank you

$$f_{3t}(x) = \frac{\log(25-(-1.8(x)))}{\log(x)}$$

edit: to clarify I am under the assumption that the inverse currently does not exist as I was trying to make a formula to solve this polynomial (x^n-1.8x=25) once I had that solved I could change up the numbers to eventually find a formula for (x^n+ax) I had worked on this for about 3 weeks and while I did manage to obtain the inverse to n I need the inverse of that equation to solve for it hence the original question I have only lightly skimmed through your answers but I will go back and check some of the links you provided me hopefully together we can find the inverse formula for this equation I also didn’t mention that it was currently unsolved as 1 I didn’t know if that’s true and 2 I didn’t want to discourage anyone from trying hopefully that answers the reason behind the question if you have anything else you want me to explain like the work I did to get that formula or other feel free to let me know! P.S sorry for the lack of punctuation I’m just a bit excited I can finally have others to have a look at it.

edit#2: here is what the inverse should look like image of the inverse and here was the closest I got to finding it my attempts to approximate the inverse hope all of this helps!

edit#3: here is all the work I’ve done to get the formula sorry in advance for all the disorganization https://www.desmos.com/calculator/slkoghiext https://www.desmos.com/calculator/t4qt5rvsp5 https://www.desmos.com/calculator/tvxtd6fvzu

https://www.desmos.com/calculator/gnrfbz0awz

https://www.desmos.com/calculator/9lpaor8brm there’s more but those are either not important or completely failed as stated before hope this helps!

$\endgroup$
13
  • 3
    $\begingroup$ I'm quite sure that function does not have a closed form algebraic solution. I think you will need numerical methods. If you edit the question to tell us where the function comes from and why you are inverting it we might be able to help. $\endgroup$ Commented Jun 9, 2022 at 13:15
  • $\begingroup$ This might be useful. $\endgroup$ Commented Jun 9, 2022 at 13:23
  • $\begingroup$ Ah this is referring to the lambert W thingy I’ve heard about it and tried to use it but I ultimately confused myself when trying to implement it to the formula which was my final attempt at doing it mostly on my own so I decided to come here and share the information to see if anyone else could figure it out. $\endgroup$ Commented Jun 9, 2022 at 16:03
  • $\begingroup$ There is no formula for the solution to the equation $x^n-1.8x=25$ for general $n$. I don't think many people here will look at all your desmos attempts. $\endgroup$ Commented Jun 9, 2022 at 16:06
  • $\begingroup$ @EthanBolker I understand that but seeing how I was able to get the inverse of the formula for the solution I am now confident that it does exist maybe in the form of a summation or exponential point being I don’t wanna give up on it just yet but thanks for confirming that there is no formula to solve this as of yet! $\endgroup$ Commented Jun 9, 2022 at 16:09

1 Answer 1

0
$\begingroup$

According to this answer, there is no definite inverse. Are you asking for an approximation? What is the specific question of the, say, homework, or did you think of this question up? You might helpful sources under if you search log. Usually writing "basically" as the reason of why you want to solve a problem isn't enough or even useful. (Also note that I am not smart enough to come up with an approximation, as I am only a seventh-grader).

Your function $f$ is an open bijection, $f^{-1}$ exists, but does not have an easy analytical expression.

$\endgroup$
2
  • 2
    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Jun 9, 2022 at 14:35
  • $\begingroup$ Btw I’m very new to posting questions so please excuse me for any bad formatting or mistakes I’ll try to fix them as I go so feel free to point out anything I might of missed thanks! $\endgroup$ Commented Jun 9, 2022 at 15:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .