What is the range and domain of a logarithm function?

[d_leet]

Actually there's still something I'm not getting, on another forum someone is saying that when you consider that log_b (x) and b ^ x are inverses, there should be an immediate realization of the problem, anyone able to help me understand this exactly?

The definition of inverse functions has been posted several times already but I will do it again.

If two functions f(x) and g(x) are inverses of each other then

f(g(x)) = g(f(x)) = x

Do you understand this?

Thus since log_b(x) and b^x are inverse functions if we let f(x) = b^x and g(x) = log_b(x)

Then f(g(x)) = b^log_b(x)

And since we said that these functions are inverses then f(g(x)) = x which means that b^log_b(x) = x.

 

[Byrgg]

"Thus since log_b(x) and b ^ x are inverse functions if we let f(x) = b ^ x and g(x) = log_b(x)

Then f(g(x)) = b ^ log_b(x)"

How did you get from the first line to the second in this? I'm sorry, I'm just not seeing how you did it.

I know that f(g(x)) = g(f(x)) = x, but I got confused when you did the above...

 

[d_leet]

"Thus since log_b(x) and b ^ x are inverse functions if we let f(x) = b ^ x and g(x) = log_b(x)

Then f(g(x)) = b ^ log_b(x)"

How did you get from the first line to the second in this? I'm sorry, I'm just not seeing how you did it.

I know that f(g(x)) = g(f(x)) = x, but I got confused when you did the above...

The log function is DEFINED as the inverse of the exponential function.

If we have a function y=f(x)

then there is a function x = g(y), such that the function y=g(x) is the inverse of f

Now let's look at this in terms of the exponential and logarithmic functions.

we have y= b^x
Now we want to define a function x = g(y) that will be the inverse of this function, and we define this function to be the logarithm so that if

y = b^x
then
x = log_b(y) these two equations are the exact same thing, and so y = log_b(x) is the inverse of y = b^x.

Again letting f(x) = b^x
and g(x) = log_b(x)

so when I take f(g(x)) all that I am doing is taking teh original x in f(x) and putting g(x) there instead

f(x) = b^x
f(g(x)) = b^g(x)
f(g(x)) = b^log_b(x) = x

Is this helping at all?

 

[Byrgg]

Ok, I think I've got it now:

let f(x) = b ^ x and g(x) = log_b(x)

f(x) = b ^ x

and x = f(g(x))



therefore subbing all the x's with g(x) brings the folllowing about:

f(g(x)) = b ^ g(x) = x
but g(x) = log_b(x)
therefore f(g(x)) = b ^ log_b(x) = x

Yes I think I finally got it all, did that look right? It's basically just what you did(I'm pretty sure) but I sort of put it into my own words.