A question about logarithm definitions

[zid]

Hi,

I've started doing logarithms recently but I got a problem understanding the definitions.

It says that in log[a](b)=n a and b can't be negative, and naturally n can't be either that way. But I don't understand why?

If log[a](b)=n is a^n=b, then why can't a and b be negative numbers? For example let's say a=-3 and n=3, then b would be =-27. Why can't a log like this exist then: log[-3](-27)=3. Everything is still in real numbers, there are no complex ones, so what's the problem? I can easily tell that log[-3](-27)=3 and that 3 is the only solution to this function. I can understand that logs like log[3](-27) go into complex realms, but why logs like log[-3](9)=2 which obviously remain in the realms of real numbers.

Can someone explain this a bit more detailed to me please? I don't want to continue doing log exercises without understanding why I have to write a>0 and b>0 every time I do an exercise. Also, I read somewhere already that it has to do something with logarithms being an inverse function of exponential functions and because exponential functions are always positive, logarithms also have to be. But I don't get it at all.

Please help me out.

 

[galactus]

Let's use an example with base 10. That is the most common.

\(\displaystyle log_{10}(100)=2\)

Because \(\displaystyle 10^{2}=100\)

But, suppose we had \(\displaystyle log_{10}(-100)=k\)

What does k equal?. Can \(\displaystyle 10^{k}=-100\)?. No value of k can satisfy that, unless we are in the complex realm perhaps.

 

[zid]

Let's use an example with base 10. That is the most common.

\(\displaystyle log_{10}(100)=2\)

Because \(\displaystyle 10^{2}=100\)

But, suppose we had \(\displaystyle log_{10}(-100)=k\)

What does k equal?. Can \(\displaystyle 10^{k}=-100\)?. No value of k can satisfy that, unless we are in the complex realm perhaps.

Ok.

As I said, I get that.

What about \(\displaystyle log_{-10}(100)=k\)? The solution is still in the realm of real numbers even though the base is <0. Or what about \(\displaystyle log_{-3}(-27)=k\). The solution is again in the realm of real numbers even though both base and number are <0.

 

[fasteddie65]

The reason for the restriction has to do with a property of functions called continuity. You will probably study that in your next course (pre-calculus or analysis or whatever it is called at your school). A function such as f(x) = a^x, where a < 0, is NOT continuous, which causes all kinds of problems when trying to work with it. By requiring a > 0, we do not have to deal with the pesky business of discontinuity at all.

 

[zid]

The reason for the restriction has to do with a property of functions called continuity. You will probably study that in your next course (pre-calculus or analysis or whatever it is called at your school). A function such as f(x) = a^x, where a < 0, is NOT continuous, which causes all kinds of problems when trying to work with it. By requiring a > 0, we do not have to deal with the pesky business of discontinuity at all.

Let me see if I got it right. a and b can be <0 and still give results in real numbers, but most of the time they don't, so we practically don't work with <0 just to avoid those numbers? That's why we put a>0 at exponential functions? And if we got the rule a>0 the rule b>0 automatically follows since any a>0 number on any exponent is >0?