2

Since the definition of a logarithm is :

$$y=\log_a x\hspace{0.1cm}\Rightarrow\hspace{0.1cm} x=a^y$$

Suppose we have :

$$(-8)=(-2)^3$$

Does this mean it is equivalent to:

$$\frac{\log(-8)}{\log(-2)}=3$$ ?

Nathaniel Bubis
  • 33,018
Chung Ren Khoo
  • 235
  • 1
    If you are working over the reals, no, what you have written is nonsense. If you assume that $-8$ and $-2$ are complex numbers, however, you can make sense of things (though it is non-trivial to do so). – Xander Henderson Aug 15 '17 at 03:32

1 Answers1

6

the definition of a logarithm is: $\;y=\log_a x\hspace{0.1cm}\Rightarrow\hspace{0.1cm} x=a^y$

The real logarithm is only defined for $\,x \gt 0\,$.

Suppose we have: $\;(-8)=(-2)^3$.  Does this mean it is equivalent to: $\,\frac{\log(-8)}{\log(-2)}=3$ ?

Suppose we have $\,-1=(-1)^3\,$. Does this mean that $\,\frac{\log(-1)}{\log(-1)}=3\,$?

dxiv
  • 76,497
  • 1
    Is the second part the reason why real logarithms are restricted to $x>0$?

    (Also, how do you cite part of my question? Been wondering how to do that for awhile now.)

     – Chung Ren Khoo Aug 15 '17 at 03:36
  • Suppose we have that $1=1^3$. Does this mean that $\log 1/\log 1=3$? (Your post doesn't explain that the sign of the argument isn't all that matters.) – symplectomorphic Aug 15 '17 at 03:36
  • 1
    @ChungRenKhoo I assumed that you meant the real function logarithm. The complex logarithm is a multi-valued function. The proposed equality doesn't hold as such in that context, either, but for more "complex" reasons (pardon the pun). how do you cite Start a new line with > then what follows will be displayed as a citation. To copy/paste text from an existing post into the citation (including formatting and formulas), click edit, copy-paste the source text, then cancel out of edit mode. – dxiv Aug 15 '17 at 03:42
  • 1
    @symplectomorphic The point you raise is about division-by-0 which is of course a valid point in general, but it is not particularly relevant in the context here. One has to worry that the two quantities are defined, first, before worrying that the denominator may be $0,$. Besides, the crux of the question is not about fractions, but rather whether $,\log x^3 = 3 \log x,$ holds for negative numbers. – dxiv Aug 15 '17 at 03:44