Why Extraneous Solutions Occur w/ Equations & Injective Functions

[SweatingBear]

Suppose we have two expressions, \(\displaystyle e_1\) and \(\displaystyle e_2\), and apply a function \(\displaystyle f\) to both expressions (or both sides of the equation). First

\(\displaystyle e_1 = e_2 \, ,\)

and then

\(\displaystyle f(e_1) = f(e_2) \, ,\)

The way I have understood this concept is that if you have two expressions and apply a function, you can always reverse the step as long as the function is injective (or one-to-one). This is why squaring can yield extraneous solutions because \(\displaystyle f(x) := x^2\) is not injective.

For example if we have

\(\displaystyle \ln (x - 4) = \ln (2x - 6) \, .\)

Let us apply \(\displaystyle f(x) := e^x\) (the function is injective!) and thus have

\(\displaystyle x - 4 = 2x - 6 \, .\)

But the equation we have arrived does not provide a solution to the original equation, in spite of the fact that we applied a one-to-one function.

I already understand that it can be comprehended if one thinks in terms of functions and their domains, but I am strictly speaking interested in thinking in terms of applying functions to both sides of an equation.

Why did it not work to apply the exponential function despite the function being injective?

 

[I like Serena]

$e^x$ is not defined if $x$ is not defined.
You can apply 1-1 (bijective) functions all you like, but you still have to check if the argument was defined in the first place.

Now if we extend the domains and ranges to the complex numbers, we can get some more results. Although we will still not have neat 1-1 inversions.

 

[SweatingBear]

Ok but if we for example have

\(\displaystyle x = \sinh (\theta) \, ,\)

and then apply \(\displaystyle f(z) := z^2\), it would yield

\(\displaystyle x^2 = \sinh^2 (\theta) \, .\)

Now I am worried whether \(\displaystyle x^2 = \sinh^2 (\theta) \, .\) contains some extraneous solutions or something else hidden that I should heed...

Sidenote: I am working on the indefinite integral of \(\displaystyle \sqrt{1+x^2}\).

 

[I like Serena]

Yes, you are introducing extraneous solutions, since a square introduces indeed an extra solution.
At least you don't lose solutions, so when you check your final solutions against your original equation, you're good to go.

 

[SweatingBear]

Yes, you are introducing extraneous solutions, since a square introduces indeed an extra solution.
At least you don't lose solutions, so when you check your final solutions against your original equation, you're good to go.

Alright thanks