# Why is $\log_{-7}49$ undefined instead of $2$?

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?

• Base must be positive and $\ne 1$ Jan 10, 2021 at 20:52
• 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 Jan 11, 2021 at 9:01

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.

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 $$a \leq 0$$.

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.

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.

• Welcome to Math SE! Please typeset your equations in MathJax. Jan 14, 2021 at 6:07