Correct me in case I am wrong, but I think that one of the major principles of equation solving is:
If $a = b$, then $f(a)= f(b)$, $f$ being a function.
In words "applying the same function to both sides of an equation yields the same output", or maybe "applying the same function preserves equality".
Suppose that I have to solve: $$2^x = 16.$$
I apply the $\log_2$ function on both sides, which gives:
$$\log_2(2^x) = \log_2(16).$$
But $\log_2(2^x) = x$ (by definition of $\log_2$).
Hence, $x = \log_2(16)=4$.
But I come across a problem when I try to apply the aforementioned alleged principle to quadratic equations.
Suppose I have to solve: $$x^2 = 16.$$
I first apply the square root function to both sides, which yields:
$$\sqrt{x^2} = \sqrt{16}$$
and (being given that $\sqrt{x^2} = |x|$)
$$| x | = 4.$$
But now I am stuck, so to say.
The reason is that I would need an "undoing absolute value function". But this "function" is not a function since it gives back $2$ outputs: $x$ OR $- x$.
Which forces me to say: $x = 4$ or $-x = 4$ (that is: $x = -4$).
Hence my question: is this an exception to my alleged principle, or I am wrong in assuming that the rule I stated is the major principle of equation solving?
If I am not applying the "$a=b \implies f(a) = f(b)$" principle, what I am doing formally in the last step of the equation solving process given above as example?
When I go from $| x | = 4$ to $x = 4$ OR $-x = 4$, am I applying a relation ( but no longer a function)?
Can one say "applying the same relation on both sides preserves equality?"
Any solution of the equation $a=b$ is also a solution of $f(a)=f(b)$. The converse need not be true.
In the simple case $2x=2$, the only solution is $x=1$. But the solutions of $4x^2=4$ are $1$ and $-1$.
There is no contradiction. The function we applied, namely squaring, is not one-to-one. From $a^2=b^2$ you cannot deduce $a=b$.
In the example with the logarithm, the function you apply is one-to-one, so it preserves equality in both directions: if $\log a=\log b$, then also $a=b$.
What about the square root? It is one-to-one when applied to nonnegative numbers. Indeed, you find $|x|=4$, which is solved by $x=4$ or $x=-4$, exactly as the equation $x^2=16$.