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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..)

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3 Answers 3

up vote 16 down vote accepted

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 $x,y \ge 0$.

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This is nitpicking, but for the sake of clarity, it might be good to add that $f$ must be bijective when viewed as a function from its domain to it range. In other words, $f$ must only be injective when the co-domain is arbitrary. –  brom Dec 2 '12 at 2:34
@brom: Good point, thanks! –  Clive Newstead Dec 2 '12 at 2:37

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.

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Wouldn't you need to check the answer if at least one step is not invertible, as opposed to each of them like you said? –  Thomas Dec 2 '12 at 5:39
Ah, I see what you mean. I meant, suppose we solve the equation by only using bidirectional (iff) implications, and a single forward implication at some point, then we still need to check the results by plugging them in the first equation, right? –  Thomas Dec 2 '12 at 6:12
Thanks, that makes sense! –  Thomas Dec 2 '12 at 7:02

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 $x = y$.
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Also simply multiplying by 0 will destroy an equation by then allowing all solutions. –  Mark Hurd Dec 2 '12 at 10:34

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