Taking the same thing of both sides of an equation

Question

I have questions regarding the validity of taking:

1. The same function (inverse or otherwise) on both sides of an equation. For example, is it possible to conclude $f(x)=f(y)$ from $x=y$ or $f(x)=x$ from $f^{-1}(f(x))=f^{-1}(f^2(x))$ from $f(x)=f^2(x)$? If it is valid, what are some other rules that you can conclude using this? For example, consider the functional equation $f(x)+f(\frac{x-1}{x})=x+1$; is it possible to conclude $x+\frac{x-1}{x}=f^{-1}(x+1)$?

2. Logs on both sides of an equation. I am more accustomed to this operation, but I don't understand why it is valid.

Please tell me if the above operations are legal and why.

Answer 1

You can usually apply a function to both sides of an equation as long as you are careful. Let's look at some of the examples you mentioned.

1. $X=Y$ implies $f(X)=f(Y)$. This is true! If $X$ and $Y$ are equal then applying any function to both should give the same thing. This is what people mean when they say that a function is 'well-defined'.

2. $f(X)=f^2(X)$ implies $f^{-1}(f(X))=f^{-1}(f^2(X))$ implies $X=f(X)$. This is true so long as $f^{-1}$ indeed exists as a function. For instance, the function $f(X)=X^2$ does not have a well defined inverse since $f(X)=f(-X)$ so $f^{-1}(f(X))$ could be either $X$ or $-X$.

3. $X+ \frac{X-1}{X}= f^{-1}(X+1)$ implies $f(X)+f(\frac{X-1}{X})=X+1$. Careful! this is not true. The problem is that when you apply $f$ to both sides, the left becomes $f(X+\frac{X-1}{X})$ and this need not be the same as $f(X)+f(\frac{X-1}{X})$.

4. You can take logs of both sides so long as you keep in mind that $\mathrm{log}(X)$ only makes sense for $X>0$.