I am really not sure either, but letīs say you take a negative number like -2, and raise it to 1/2. The square root of negative numbers isnīt definied for real numbers.
My text asks me to describe why it is I cannot put a negative base into a logarithm, and I can't figure why. I can see how it would cause problems, what with -2^3 won't give you a positive eight....? I don't get it. It seems to me that it could work out sometimes, but the text clearly states, "You cannot have a negative base". Thats all it says though. I reckon it assumes I should know, or it should be blatantly obvious, but I don't get it. Any thoughts?
I think it is because a logarithm is a inverse function of f(x) = a^x, where x can be any real number. But a cannot be anything. If a = -2 then it is not a proper function, you've got something that "jumps", and if x = 1/2
(-2)^(1/2) is not defined. so f(x) = a^x is defined
So we've got a function f(x) that always has positive outcome, the inverse function therefore has to start with a positive (base) number.