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Do the logarithmic rules work when taking logs of functions as opposed to numbers?

i.e. suppose $f$ is a function and $n$ is a real number, is $\log (f(x)^n) = n · \log(f(x))$?

Kenta S
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Peter_Pan
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    Of course, because the equality is true for any specific $x$. – Mark Jun 13 '19 at 21:42
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    @Mark : What you have said is meaningless. Can you elaborate? – MPW Jun 13 '19 at 21:43
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    Why is it meaningless? Put any constant $x$ and you will get an equality. Two functions are equal if they are equal at all points. – Mark Jun 13 '19 at 21:44
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    Okay, I guess I understand what you mean. Note that, unfortunately, OP has omitted any mention of $x$ on the RHS. – MPW Jun 13 '19 at 21:45

2 Answers2

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Yes it will work.

$$\log(f(x)^n) = \log(f(x)\times f(x) ... \times f(x)$$ $$= \log(f(x)) +\log(f(x)) + ... +\log(f(x))$$ $$=n \times \log(f(x))$$

Vizag
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  • Wouldn't this only work for $f(x) > 0$? Especially if $n$ is even. Suppose $f(x)=-2$, then $f(x)^n>0$, so $\log(f(x)^n)$ is defined, but $\log(f(x))$ isn't. – JAD Jun 14 '19 at 07:16
  • @JAD log(-1) = π * i - i.e. log of a negative real number is a complex number, so answer is NO - it will perfectly work on negative values, except that number domain changes. And besides this isn't related to a f(x), but to log() domain in general. – Agnius Vasiliauskas Jun 14 '19 at 07:34
  • @AgniusVasiliauskas alright, thanks. – JAD Jun 14 '19 at 07:36
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Yes, of course.

Because $f(x)$ is still a number for any $x$.

It even works on expressions: for example $\log((x^2+3)^9)=9\log(x^2+3).$

Saketh Malyala
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