What operations on an equation cause it to be destroyed? I approached my calculus professor about something he said which didn't make much sense to me - He says that in the process of calculating $\lim_{x\to\infty} f(x)^{g(x)}$, you can convert it to $\lim_{x\to\infty}e^{g(x)\cdot \ln(f(x))}$. I understand that much - $e$ and $\ln$ are inverse functions, so they cancel each other out and the end result is the same equation.But when he was showing us how to do implicit differentiation, he says that $y=\ln(f(x))$ can be re-written as $e^y=e^{\ln(f(x))}$, or ultimately $e^y=f(x)$. Same goes for converting a function $y=f(x)^{g(x)}$ to $\ln(y)=g(x)\cdot \ln(f(x))$.
My first impression was to say WTF? I would have thought that raising both sides of the equation to a power of $e$ would destroy the equation... but apparently not. So my question is, what operations can you safely do like this to both sides of an equation - or if it's more concise, what list of things can you NOT do? The only two I'm aware of currently after speaking with my prof are squaring and taking the square root (or cubing, etc..)
 A: We start with an equation (1) and try to rewrite it into equation (2). Now it can be the case that (1) implies (2) and (2) implies (1) so that they have the same solutions, or that (1) implies (2) only so that any solution of (1) also satisfies (2) but a solution of (2) does not necessarily satisfy (1).
If $x=y$, then clearly $e^x=e^y$. The next question to ask is if the converse is true. Now if $e^x=e^y$, then clearly $\ln e^x=\ln e^y$. But we know that $\ln e^x=x$ and $\ln e^y=y$, so it follows that $x=y$.
Sometimes when we solve equations we let the direction of implication go one way only and sometimes both. We just need to be aware of what we are doing. If the direction is only forward for each of the steps, then we need to plug in the solution to the final equation into the first one to make sure it really works.
A: One operation that will destroy an equation is dividing both sides of the equation by zero! (It makes no sense to equate undefined with undefined!)
Seriously, sticking with defined operations and manipulations on expressions: 
Manipulating each side of the equation in the identical manner maintains "an" equality, but not necessarily the equality.
Manipulations sometime fail to preserve all the information of the original equation (like all potential solutions). 
And at other times such manipulations, in fact, introduce solutions to the manipulated equation that are not solutions to the original equation.


*

*E.g., If $x = y$, then $x^2 = y^2$: this is a true statement. However $(-2, 2)$ solves $x^2 = y^2$, but is not a solution to $x = y$.

A: A re-formulation of your question is as follows:

Under what conditions on $f$ is it valid to say
  $$x = y\ \ \text{if and only if}\ \ f(x)=f(y)$$
  ?

The answer is then quite simply that we need $f$ to be injective. Equivalently, we need $f$ to have a left-inverse. And, obviously, $x$ and $y$ must be in the domain of $f$.
The reason why saying $y=\ln f(x)$ is the same as saying $e^y=e^{\ln f(x)}$ is precisely because the function $x \mapsto e^x$ has a (left-)inverse; and that inverse is $y \mapsto \log y$ since $\log e^x=x$.
The reason why it's not true to say things like $-2 = 2 \Leftrightarrow (-2)^2=2^2$ (which is clearly an invalid statement) is because $x \mapsto x^2$ is not injective. However, it is injective (in fact, bijective) when you restrict to positive integers, and so the statement $x=y \Leftrightarrow x^2=y^2$ is valid when $x,y \ge 0$.
