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let $f(x) = \log{x} + \log{(x - 2)}$

let $g(x) = \log{(x^{2} - 2x)}$

if

$f(x) = g(x)$

then why in the RHS, x can be -1 and in the LHS not ? There's a restriction that i do not know to when its valid to expand the logarithm of a product ?

displayName
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    What is $\log{x^{2} - 2x}$ ? Do you want $\log(x^{2} - 2x)$ or $\log(x^{2}) - 2x$ or $(\log x)^{2} - 2x$? – 311411 Oct 02 '21 at 00:59
  • When we're talking about the real numbers, the argument to $\log$ can only be positive (things are a bit more complicated in complex analysis). If $x$ is negative, then $x^2 - 2x$ is positive, but obviously $x$ and $x-2$ aren't. – Connor Harris Oct 02 '21 at 01:32
  • so $f(x) \neq g(x)$ ? – displayName Oct 02 '21 at 01:45
  • In Real Analysis, consider the two functions $f(x) = x$ and $g(x) = [\sqrt{x}]^2.$ For $x \geq 0, f(x) = g(x).$ For $x < 0, f(x)$ is defined and $g(x)$ is undefined. – user2661923 Oct 02 '21 at 02:23

2 Answers2

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This is easier to see if you plug in the actual number for both functions.

$$g(-1) = \log((-1)^2-2(-1)) = \log(1+2) = \log(3)$$

$$f(-1) = \log(-1) + \log(-3)$$

Now notice that $-1*-3 = 3$. So $\log(3) = \log(-1*(-3))$. But you can't actually split the product into a sum of two logarithms here, because then you'd be putting invalid numbers into the logarithm (if you want the result to be real).

Basically, $\log(ab) = \log(a) + \log(b)$ iff $a$ and $b$ are both positive if we're talking about real number valued logarithms. Which means that $f(x) = g(x)$ only if $x-2 > 0$, i.e. $x>2$.

However, for the part of $g(x)$ where $x < 0$, you can extend $f(x)$ by defining it as equal to $g(x)$ in that region.

Breaking Bioinformatics
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In terms of functions, the Domain of $f(x) ≠ g(x)$. Never simplify graphical functions unless the question specifies it. One can relate the two functions, $Range_f \in Range_g$ and $Domain_f \in Domain_g$

You can see this through their graphs:

$f:$ f(x)

$g:$

g(x)

Its the same with: $\frac{x^2}{x} ≠ x$ for $x = 0$.

D[$\frac{x^2}{x}$] = $\mathbb R$ \ {$0$}, however D[$x$] = $\mathbb R$.

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