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#### sweatingbear

##### Member

- May 3, 2013

- 91

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

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?