Applying any injective function d(x) to the equation f(x)=g(x) is always an equivalent transformation? May it lead to alien or missing roots? Say I have an equation (1): $f(x)=g(x)$
Then I apply an injective function to both sides of (1). The resulting equation is guaranteed to have same roots as (1) and only such roots as (1)? No missing or alien roots possible?
For example I raise both parts to the third power, or make (1) to be $e^f(x)=e^g(x)$ because $y=x^3$ and $y=e^x$ are injective for any $x$ from real.
Or even say I care only for the roots of (1) from the interval $x=[a, b]$. Then I can apply a function which is injective on that interval, say $y=x^2$ if I'm interested only in positive roots of (1).
Is my reasoning correct everywhere?
Summary (of answers):
Below algorithm is totally complete/correct (inspired by Wouter answer):
Firstly, if domain of d(x) is x=R and d(x) is injective on all its domain, then I can safely apply d(x) to both sides of (1) and this is absolutely equivalent transformation.
Secondly, if domain of d(x) is not x=R, or domain of d(x) is x=R but d(x) is injective only within x from some subset from d(x) domain, then the following holds:
if I care about solutions of the original equation $f(x)=g(x)$ (1) only from the interval $x=[a;b]$, then I absolutely need to check (only) TWO things first - that the function $d(x)$ that I want to apply to both parts of the equation (1) is:


*

*defined in Range_of_f(x) and Range_of_g(x) (in other words that Domain of d(x) includes union of Range_of_f(x) and Range_of_g(x)) AND

*d(x) is injective on the interval of x being the union of Range_of_f(x) and Range_of_g(x) - 


Given ranges for $f(x)$, $g(x)$ in steps 1 and 2 above are calculated for $x=[a;b]$
If both 1 and 2 are satisfied, then I just apply d(x) to both parts of equation (1), solve it and getting roots, if they are from $x=[a;b]$ then these (and only these) are roots of (1) on $x=[a;b]$ interval (no missed, no alien roots on that interval).
P.S.  Aren't there any issues with zeros of d(x) function which is applied to both sides of equation $f(x)=g(x)$ ?
 A: Let $x \in X$, $f(x)=g(x)$ and let $d$ be an injective function. Then $d(f(x)) = d(g(x))$, so x is still a solution to the equation (no missing solutions)
Similarly, let x be such that $d(f(x)) = d(g(x))$. Since $d$ is injective, $f(x)=g(x)$ i.e: x is a solution to the original equation (no intruders)
A: You are correct, provided the domain of the injective function contains the codomain of both $f$ and $g$.
For example
$$x=-3$$
You don't want to do
$$\log(x)=\log(-3)$$
because although $\log(x)$ is injective, it is only injective in $\mathbb{R}^+$, and $-3\not\in \mathbb{R}^+$
A: Suppose $f : X \to Y$ and $g : X \to Y$ are two functions and $h : Y \to Z$ is an injective function. Define $A = \{x \in X : f(x) = g(x)\}$ and $B = \{x \in X : h(f(x)) = h(g(x))\}$. Then $A = B$. 
Proof
If $x \in A$, then $f(x) = g(x)$. Thus, $h(f(x)) = h(g(x))$. Therefore, $x \in B$. This shows $A \subset B$. Note that this would have been true even if $h$ weren't injective. 
If $x \in B$, then $h(f(x)) = h(g(x))$. Since $h$ is injective, this means $f(x) = g(x)$. Therefore, $x \in A$. This shows $B \subset A$. Note that injectivity of $h$ was necessary for this step.
Since $A \subset B$ and $B \subset A$, we may conclude $A = B$. That is, the solutions of $f(x) = g(x)$ and $h(f(x)) = h(g(x))$ are identical. 
