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When you have $\log_{7}49$, you can tell that it is equal to $2$ because $7^2 = 49$. But, when you have $\log_{-7}49$, isn't this also equal to $2$ because $(-7)^2 = 49$ also? Instead, this is undefined.

Can someone explain why this is?

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    $\begingroup$ Base must be positive and $\ne 1$ $\endgroup$
    – Raffaele
    Jan 10, 2021 at 20:52
  • $\begingroup$ What would say is $\log$ $7$ base $-7$ ? It is not $1$, but with the usual properties of logarithms $\log$ $7^2$ base $-7$ should be two times that $\endgroup$
    – Henry
    Jan 11, 2021 at 9:01

4 Answers 4

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In general it is not possible to take a number $x$ and a negative number $n$ and find an exponent $y$ such that $n^y=x.$ For example, it won't work if $x=7$ and $n=-7.$ It works out in the specific case of $x=49$ and $n=-7$, but it's just not useful enough to define the logarithm function for a few special values where logarithm by a negative number actually works.

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When working in the real numbers, we require that the base of the logarithm is greater than $0$ (and not equal to $1$). While the expression $$ \log_{-7}(49)=2 $$ does have some sense to it, it is formally incorrect for this simple reason.

This begs the question—why do we avoid negative bases? This is because the function $\log_{-7}$ would be ugly. It would only be defined for a limited subset of the real numbers, and it wouldn't be differentiable, unlike other logarithm functions. The exponential function $(-7)^x$ only makes sense for some rational $x$ (where the denominator is odd), and so exponentiation of negative numbers is a touchy subject in general. Note that the definition of $a^x$ is $$ \exp(x\log(a)) $$ where $\log$ is the natural logarithm and $\exp$ is the exponential function ($\exp(x)=e^x$). Negative bases have to be dealt with separately, as $\log(a)$ is undefined for $a \leq 0$.

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If you consider the function $f(x) = b^x$, the function logarithm in base $b$ is defined to be the inverse of $f$. In other words, if we denote $f^{-1}$ this function, we should have $$(f \circ f^{-1})(x) = x \quad \text{and} \quad (f^{-1} \circ f)(x) = x.$$ However the notion of inverse function make sens only for bijective functions. Since $b^x$ in not bijective for $b = 0$ and not even well-defined for $b < 0$ (indeed, for $b = -1$ and $x = 1/2$ for example, $b^x$ is not a real number), the inverse function is only defined for $b > 0$ and we note $\log_b$ this function.

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Log49 to the base -7 is not a real number. However it is a complex number

Log49 to the base -7= ln49/ln(-7)

Ln(-7) is not a real number Remember exp(ia)= cos(a)+isin(a)

-1= exp(iπ) ln(-7)= ln(-1)+ln(7)=ln(exp(iπ))+ln(7) =Ln(7)+ iπ. (Complex number)

Log49 to the base -7= log(49)/( some complex number) Hence proved.

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  • $\begingroup$ Welcome to Math SE! Please typeset your equations in MathJax. $\endgroup$
    – KingLogic
    Jan 14, 2021 at 6:07

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