$\log _{-3}\left( 9 \right)$?

Question

Can you have a negative base for a logarithm? If the answer is no, why can you not? Isn't the answer to the above just $2$?

Similarly, is $-3$ also a solution to the equation $\log _{x}\left( 9 \right)\; =\; 2$ (I know one solution is obviously 3).

And what would be $\log_{-3}(8)$ equal to? – Peter Košinár Dec 20 '13 at 19:01 — Peter Košinár, Dec 20 '13 at 19:01

score 5 · Answer 1 · answered Dec 20 '13 at 19:01

5

By definition for $a>0$ we have $$\log_a(x)=\frac{\log_e(x)}{\log_e(a)}$$ where $$\log_e(x)=\int_1^x\frac{dt}{t}$$

answered Dec 20 '13 at 19:01

Short but to the point: nails it! +1 – amWhy Dec 21 '13 at 17:50
2

Congratulations!! $\Large > 30\text K$!!! – amWhy Dec 21 '13 at 17:52

score 1 · Accepted Answer · answered Dec 20 '13 at 19:40

1

By definition $\log_{-3}(x)$ would be the $y$ such that $(-3)^y=x$. The trouble here is that $(-3)^y$ is not well-defined unless $y$ is an integer, so you won't be able to solve this for very many different $x$.

Even if you say that $(-3)^y$ is meaningful whenever $y$ is rational with odd denominator, this only gives you countably many different $x$ for which $\log_{-3}(x)$ can exist, so it won't behave very much like ordinary logarithms. (The corresponding antilogarithm wouldn't be continuous, for example).

answered Dec 20 '13 at 19:40

hmakholm left over Monica

286,031

Logarithms with negative bases or negatives arguments are always complex functions. $log_{-3}(9) = { {log(9)} \over {log(3) + \pi i}} = {{log 3 log 9 - \pi i log 9} \over {log 3 log 3 + \pi \pi}}$ – Arucard Dec 24 '13 at 18:43
@Arucard: The trouble with that is $-3$ have many different complex logarithms, and each of those gives rise to a different $\log_{-3}$ function, even before we begin to consider the branch point of the logarithm in the denominator. – hmakholm left over Monica Dec 25 '13 at 03:34

$\log _{-3}\left( 9 \right)$?

2 Answers2