Awkard question with logarithms

[Byrgg]

I posted a thread about proving the following:

$b^(log_b x) = x$

a while back, and received help in understanding multiple ways to prove it. But I also asked a math teacher prior to requesting help here, he suggested 2 methods:

Substitution(also suggested by my physics teacher):

Let $log_b x = y$, which is the same as writing $b^y = x$, right? but we already said y = $log_b x = y$, therefore, by substituting we get: $b^(log_b x) = x$

That method was simple enough, I was also shown by the people here that by applying that $log_b x$ and $b^x$ were inverses, you could achieve the same final result:

let $log_b x = f(x)$, and $b^x = g(x)$, by composing the inverses you get:

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

And after much work, I finally understood this method as well.

But another way the math teacher suggested, was what he called the 'intuitive' way, simply by 'looking' at the problem, you could see the relationship to be true... my guess is that he meant you could see the first way mentioned here simply by 'looking'. Does anyone have an idea of what he may have been suggesting? Thanks in advance.

 

[Byrgg]

Woops sorry, the imaging didn't work out so well, look at the equation I was trying to prove as b^(log_b x) = x.

 

[Byrgg]

Argh, the message editor isn't working well, also, when I composed the inverses read as b^(f(x)) (with f(x) being entierely in the exponent).

 

[Byrgg]

I'll try and clear up any misconceptions here with my poor latex skills, just to go over everything...

I'm trying to prove x = b^(log_b x), which I know how to do, but I'm just curious as to what the 'intuitive' way mentioned by the math teacher is...

When I composed the inverses, it should be read as:

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

Someone please help soon...

 

[nrqed]

I posted a thread about proving the following:

$b^(log_b x) = x$

I guess you meant $b^{log_b (x)} = x$ (you must use curly braces).

It's actually "obvious" by the very definition of a log in base b. That's surely what your prof had in mind.

By definition, what does log_b(x) represent? Using only words? It represents the exponent to which b must be raised to give x! Right?

Therefore if you calculate $b^{log_b(x)}$ you of course get back x!

You see what I mean?

patrick

 

[arunbg]

I think you mean
$$b^{log_b(x)}=x$$

Think about the basic definition of logarithm. How are logarithmic and exponential functions related ?

Edit: Lately my typing has been on the slower side.

 

[nrqed]

I think you mean
$$b^{log_b(x)}=x$$

Think about the basic definition of logarithm. How are logarithmic and exponential functions related ?

Edit: Lately my typing has been on the slower side.

:rofl: :rofl:

It looks as if we are "chasing" each other around the boards! :tongue2:
And we are posting the same comments! (hey, maybe we are twins from different universes, "a la Many World Interpretation" of QM!

I am logging off now so I won't annoy you by "cutting" you off!


Best regards

Patrick

 

[Byrgg]

Yes, thanks for providing that code reference for me, $$b^{log_b(x)}=x$$ indeed that is what I was trying to write.

Ok, so let's see here, the exponent to which must be raised to reach x, this is the same as log_b(x), right? So by rasing b to this exponent, you must get x? Right?

Wow, I feel like an idiot now >_>, I'm pretty sure that's exactly what the math teacher said too, I've got to pay more attention, it's hard when you're stressing about exams and the like though, thanks for your help you two.

If my little summary there was right, then I don't think there's any more further questions.

 

[Byrgg]

Oh, also I guess it's worth bringing back(from my old thread), about
log_b (x) and b^x being inverses...

It was said that the equation $$b^{log_b(x)}=x$$ was precisely saying log_b (x) and b^x were inverses...

You would prove that statement by using the method I showed before right? Or is there something else I forgot?

Here's the method...

let y = $log_b (x)$, and y = $b^x$ by composing the inverses you get:

g(f(x)) = x = $b^{log_b(x)}$

Right?

Basically, I'm curious as to why exactly you say they are inverses in the first place, is it because of the definition?

 

[0rthodontist]

You can just take the log base b of both sides. That will save you the trouble of thinking about it, unless you're trying to work from first principles.

 

[Byrgg]

Yeah I already know that way, and yes, I was trying to work from first principles, sort of... Was I right, or does something need correcting?

 

[Hootenanny]

Basically, I'm curious as to why exactly you say they are inverses in the first place, is it because of the definition?

Basically yes, they are inverses because they are defined as such. This is like asking why is arcsine the inverse of sine.

 

[Byrgg]

Ok, and from that knowledge, you can use the method I showed(and learned here in the first place) to prove the equation(composing the inverses)?

 

[arunbg]

Yes Byrgg, I think you have got it .
Also it is worthwhile to note that
$$log_b(b^x)=x$$
This highlights the property of invertible functions.

 

[Byrgg]

Yeah I noticed that too, and I guess it's also worth mentioning how easy it is to see the compostion of inverses here...

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

so obviously those two relationships simply show the subbing of inverses into each other, which also equals x, right?

 

[Hootenanny]

Yeah I noticed that too, and I guess it's also worth mentioning how easy it is to see the compostion of inverses here...

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

so obviously those two relationships simply show the subbing of inverses into each other, which also equals x, right?

Yes, they are both functions of x and inverses of each other, so if one forms a composite function, either fg(x) or gf(x), then the result in this case simply x.

 

[nrqed]

Yeah I noticed that too, and I guess it's also worth mentioning how easy it is to see the compostion of inverses here...

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

so obviously those two relationships simply show the subbing of inverses into each other, which also equals x, right?

Yes.

If we use the notation $f^{-1}$ for the inverse of a function, then, by definition,
$$f ( f^{-1}(x)) = f^{-1}(f(x)) = x$$

It's actually a bit tricky because one must be very careful about the domains and images of the function and inverse functions and that may mess up things sometimes (for example, $\ln(e^{-2}) =-2$ but $e^{\ln(-2)}$ is clearly not defined. Or taking an inverse sin has multiple solutions).

 

[Byrgg]

I don't know, the one thing I think that still bugs me is exactly how you can look at $b^{log_b (x)} = x$ and simply automatically say that $log_b$ and $b^x$ are inverses... Since you can't really tell what the exponent of b was prior to substitution...

 

[arildno]

An exponential function is a strictly increasing function. With that, you may prove that there exists an inverse function.

 

[Hootenanny]

Let me ask you a question how can you just look at ArcCos(Cos(x)) = x and simply automatically say that ArcCos and Cos are inverses?

 

[Byrgg]

You can say those are inverses because they do opposite things to x, thus cancelling out I guess...

 

[Hootenanny]

You can say those are inverses because they do opposite things to x, thus cancelling out I guess...

Yup, what's the difference between my question and yours?

 

[Byrgg]

Hmmm, I think I see your point... b, raised to an exponent, gives x... this could be seen as a function of x I think, right?

 

[Hootenanny]

Hmmm, I think I see your point... b, raised to an exponent, gives x... this could be seen as a function of x I think, right?

It is indeed a function of x. f(x) = b^x and f^-1(x) = log_bx

 

[Byrgg]

Err, if b raised to an exponent gives x, then wouldn't it be something like b^y = x? You kind of put it the other way around there I think...

 

[Hootenanny]

Err, if b raised to an exponent gives x, then wouldn't it be something like b^y = x? You kind of put it the other way around there I think...

I was simply stating that the two functions I gave about were inverses. I'm not quite sure where your going with the b^y = x thing.

 

[Byrgg]

Well, the b^y = x thing could also be seen written as log_b (x)... which is what I was trying to say earlier... b to an exponent equals x.

 

[Hootenanny]

So what your saying is that b^y = x and that y = log_b|x| ?

 

[HallsofIvy]

So what your saying is that b^y = x and that y = log_b|x| ?

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.

By the way, Byrgg, the whole point is founded on how these functions are defined, a point made earlier. Often f(x)= b^x is defined by defining bⁿ for positive integer n and then expanding to other numbers systems in a way that keeps the "laws of exponents" true. g(x)= log_b(x) is then defined as the inverse function to f.

Conversely, it is sometimes done, in calculus, to define ln(x) as
$$\int_0^x \frac{1}{t}dt$$
show all the properties of logarithm from there and then define e^x as the inverse of that.

Of course, you could use both definitions and show that each is the inverse of the other.

 

[Byrgg]

Sorry about that, I was away for a while... had some time to clear my head a bit...

Anyway... referring to Hootenanny's example:

ArcCos(Cos(x)) = x

I realized that in this case, you can say they are inverses, because normally cos(x) and arccos(x) are functions of x, correct? In order to obtain x from cos(x), you would have to plug the inverse of cos(x) into it, or plug cos(x) into the inverse, thus obtaining the equation ArcCos(Cos(x)) = x, which is also equal to Cos(ArcCos(x)) = x, this sound ok so far?

For the function log_b (x), in order to obtain x from this, you must similarily, plug this into it's inverse, or it's inverse into it. The inverse being b^x, as said, you can do this either way, by subbing it into the inverse, or vice versa, and thus obtaining b^log_b (x) or log_b b^x. Therefore by looking at the equation $b^{log_b (x)} = x$, you can see the relationship, you can see log_b(x), and on the other side this has been turned into x, therefore, you can assume it's been plugged into it's inverse, or it's inverse has been plugged into it, right? Thus obtaining the two possible equations that I already mentioned, and am too lazy to write out again. And of course, the inverse here has to be b^x, because you are subbing the function into each x value.

Whew! That sound ok? Did I miss something, or ddoes that all sound logical. I know it's very obvious... but you know, with all my questions, it's veyr difficult to keep a clear mind.

 

[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.