Solving for the Inverse of log (x)/3

AI Thread Summary
To find the inverse of log(x)/3, it's essential to clarify the function's form, as it can be interpreted in two ways: f(x) = log(x/3) or f(x) = (log x)/3. The correct interpretation appears to be f(x) = (log x)/3, leading to the inverse function f-1(x) = b3^x. The discussion also touches on the potential confusion arising from the notation used in a biostatistics article, which may have referred to a different function altogether. Additionally, the possibility of taking the inverse of the logistic regression equation is mentioned, highlighting the complexity involved in such calculations. Understanding these nuances is crucial for accurately solving the inverse function.
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Hi

How do I take the inverse of log (x)/3? If it is just log (x), it seems quite easy to do but I don't know what to do with the division by 3.

I saw this equation in a biostatistics article and I just can't understand how to solve it. It's been so long since I did inverse functions and I would really appreciate your help.
 
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Is the function f(x) = log (x/3) or f(x) = (log x)/3 ? First, solve for the independent variable:

f(x)=\log_b (x/3) \Rightarrow b^{f(x)}=b^{\log_b (x/3)} \Rightarrow b^{f(x)}=x/3 \Rightarrow 3b^{f(x)}=x

And to find the inverse function, switch the independent and dependent variables: f-1(x) = 3bx. Through a similar process, if you have f(x) = (log x)/3, the inverse would be f-1(x) = b3x, I think.
 
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Hi pbandjay

Thanks so much for your help!

I think f(x) is (log x)/3. It is a bit confusing the way they wrote it in the article.

they had it written out like this:

f-1 {log (x)/c} where c is some constant
 
f^{-1}(\frac{log(x)}{c})

that appears to be asking for f-1 OF log(x)/c for some other function f, not for the inverse function of log(x)/c.
 
Thanks HallsofIvy

if that's the case, they were referring to the logistic regression equation. Is it possible to take the inverse of the logistic regression equation?
 
Perhaps they wrote it that way because talking about the inverse is easy, while computing a formula for it is complicated and not useful in the discussion.
 
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