Awkard question with logarithms

[Hootenanny]

Who were you responding to? I surely wouldn't say that- I would maintain that the function f(y)= b^y is only defined for positive and therefore, x= b^y must be positive.

For b positive, the function f(x)= b^x has domain all real numbers and range all positive real numbers and so log_b(x) has domain all positive real numbers and range all real numbers.

That's what I thought Byrgg was saying in his post. I wanted to make sure, before I launched into an explanation.

Byrgg, your explanation there is correct. However, I would state as Halls of Ivy and many others have that the two functions are defined as inverses.

 

[Byrgg]

I came up with another, though very crude demonstration, someone tell me if this is ok:

(x + 1) - 1 = x

Obviously, it can be seen simply that x + 1 - 1 = x, but could you also see this as a composition of inverses? Take x + 1 and x - 1 to be inverses, and thus by substituting one into the other, you receive either
(x + 1) - 1 = x or (x - 1) + 1 = x

Very simple, but I think this also shows an example.

 

[arunbg]

Yes, I think that qualifies albeit trivial .

 

[HallsofIvy]

I came up with another, though very crude demonstration, someone tell me if this is ok:

(x + 1) - 1 = x

Obviously, it can be seen simply that x + 1 - 1 = x, but could you also see this as a composition of inverses? Take x + 1 and x - 1 to be inverses, and thus by substituting one into the other, you receive either
(x + 1) - 1 = x or (x - 1) + 1 = x

Very simple, but I think this also shows an example.

In other words, if f(x)= x+ 1 then f^-1(x)= x-1 is its inverse.
Or, in words, adding and subtracting are "inverse" operations.

 

[Byrgg]

Yeah that's what I was getting at.