Inverse Functions: Logarithms & Qualitative Conclusions

[Byrgg]

In one of my older threads, I posted the following:

log_b (n) = x if and only if b ^ x = n, where b > 0, and b is not equal to one.

It was said this defines logarithm as the inverse to exponential. I don't really see how that works here, I think it just shows how you write logarithms. Could anyone clear this up?

Also, I'm wondering something about inverses in general. Say you have a simple example like (x + 1) - 1 = x. This equation says that (x + 1), and (x - 1) are inverses, correct? Is there a way to mathematically show that? Or do you simply conclude this qualitatively? I'm thinking if it's just qualitative, you could say something like this:

If (x + 1) - 1 = x, then it must be that + 1 and - 1 do opposite things to x. Thus, you can conclude that (x + 1) and (x - 1) are inverse functions, seeing as they do opposite things to x.

Does that sound right? Is there I way to mathematically show this? I know that if you take (x + 1) and (x - 1) and sub one into the other you get either (x + 1) - 1 = x or (x - 1) + 1 = x, but can you do this backwards, by starting from (x + 1) - 1 = x or (x - 1) + 1 = x?

Thanks in advance for any help I get.

 

[Gagle The Terrible]

Intuitevely, the inverse of a function f(x), noted f^(-1)(x) is such as :
f(f^(-1)(x)) = x . The inverse function "cancels" the effect of f(x) over x.
I would use this definition to prove the inverse.
The subtle question is : Does the inverse funtion exist ?

Interesting analysis arises from such questions and a lot of theorems are devoted to such a topic.

Logarithms and exponentials function can be defined from one another as being inverses , hence Exp(Log(x)) = x

Of course,this is true only if the conditions you stated are fufilled(i.e. same basis not equal to one)

For your other question, f(x) = (x+1) has (x-1) as inverse and vice versa. As you mentionned, it "works both ways".

 

[d_leet]

In one of my older threads, I posted the following:

log_b (n) = x if and only if b ^ x = n, where b > 0, and b is not equal to one.

It was said this defines logarithm as the inverse to exponential. I don't really see how that works here, I think it just shows how you write logarithms. Could anyone clear this up?

This is just how the logarithm is defined. It is defined to be the inverse function of the exponential function.As for inverse functions in general, two functions let's call them f(x) and g(x) if and only if

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

If you can show that this relationship holds for two given functions then these functions are inverses of each other.

 

[arildno]

Byrgg:
Since it can be proven that:
1. The exponential function is strictly increasing
and
2. Every strictly increasing function HAS an inverse

I really don't see your trouble.

 

[0rthodontist]

I think I recall from my high school calculus book that log x is defined as \int_1^x \frac{dy}{y}, and that e^x is defined as the inverse of log x, from which general exponentiation (exponentiation to irrational powers) is defined. I don't know how accurate this is.

 

[HallsofIvy]

In one of my older threads, I posted the following:

log_b (n) = x if and only if b ^ x = n, where b > 0, and b is not equal to one.

It was said this defines logarithm as the inverse to exponential. I don't really see how that works here, I think it just shows how you write logarithms. Could anyone clear this up?

No, "showing how you write logarithms" wouldn't have anything to do with exponentials, would it? Saying "log_bn= x if and only if b^x= n" is exactly saying that they are inverses. If f(x)= y, then f^-1(y)= x. That's exactly logb[/sup](x)= y, b^y= x. Another characterization of inverse functions is: f(f^-1(x))= x and f^-1(f(x))= x. From your equations log_b(b^x)= log_b(n)= x and b^{log_b n}= b^x= n.

Also, I'm wondering something about inverses in general. Say you have a simple example like (x + 1) - 1 = x. This equation says that (x + 1), and (x - 1) are inverses, correct? Is there a way to mathematically show that?

Yes, that is precisely f(f^-1(x))= x. Here f^-1(x)= x+ 1 and f(x)= x-1. f(f^-1(x))= f(x+1)= (x+1)-1= x and f(f^-1)(x))= f(x-1)= (x-1)+ 1= x.

Or do you simply conclude this qualitatively? I'm thinking if it's just qualitative, you could say something like this:

If (x + 1) - 1 = x, then it must be that + 1 and - 1 do opposite things to x. Thus, you can conclude that (x + 1) and (x - 1) are inverse functions, seeing as they do opposite things to x.

Yes, that's a very good, less technical, way of thinking about inverse functions: whatever f "does", f^-1 "undoes".

Does that sound right? Is there I way to mathematically show this? I know that if you take (x + 1) and (x - 1) and sub one into the other you get either (x + 1) - 1 = x or (x - 1) + 1 = x, but can you do this backwards, by starting from (x + 1) - 1 = x or (x - 1) + 1 = x?

Thanks in advance for any help I get.

Well, once again, f(f^-1(x))= x and f^-1(f(x))= x are the ways to "mathematically show this". Since applying first one and then the other, whatever one "does" to x, the other "undoes".

You can see why inverse functions are so important: in order to solve the equation f(x)= b, you must "back out" and "undo" whatever f has "done" to x. If you know the inverse function of f, you only need to apply that function to both sides of the equation: f^-1(f(x))= x= f^-1(b).

Of course, solving equations, in general, isn't that easy! That's because most functions don't have inverses. If f(x₁) and f(x₂) both equal y, what is f^-1(y)? Is it x₁ or x₂? Since there is no way to answer that, only "one to one" functions have inverse. Further, if there is no x such that f(x)= y, then there can be no f^-1(y). Only "one to one" and "onto" functions have inverses. Since (2)²= 4 and (-2)²= 4, as simple a function as f(x)= x² does not have an inverse. f(x)= x² is neither "one to one" nor "onto". Instead we "modify" the function to : g(x)= x² if x\ge 0 and is undefined for x< 0. We can say that f is "one to one" and "onto" the set of non-negative numbers. Now, if y is a non-negative number, g^{-1}(y)= \sqrt{y} is defined. Of course, if we have the equation f(x)= x²= -4, precisely because f(x)= x² is not "onto" the set of all real numbers, we have to say there is no solution because f^-1(-4) is not defined. Also, because f(x)= x² is not "one to one", to solve f(x)= x²= 4, we have to re-interpret the equation as g(x)= 4 to get x= 2, then recognize that there are two solutions, 2 and -2.

 

[HallsofIvy]

In one of my older threads, I posted the following:

log_b (n) = x if and only if b ^ x = n, where b > 0, and b is not equal to one.

It was said this defines logarithm as the inverse to exponential. I don't really see how that works here, I think it just shows how you write logarithms. Could anyone clear this up?

No, "showing how you write logarithms" wouldn't have anything to do with exponentials, would it? Saying "log_bn= x if and only if b^x= n" is exactly saying that they are inverses. If f(x)= y, then f^-1(y)= x. That's exactly logb[/sup](x)= y, b^y= x. Another characterization of inverse functions is: f(f^-1(x))= x and f^-1(f(x))= x. From your equations log_b(b^x)= log_b(n)= x and b^{log_b n}= b^x= n.

Also, I'm wondering something about inverses in general. Say you have a simple example like (x + 1) - 1 = x. This equation says that (x + 1), and (x - 1) are inverses, correct? Is there a way to mathematically show that?

Yes, that is precisely f(f^-1(x))= x. Here f^-1(x)= x+ 1 and f(x)= x-1. f(f^-1(x))= f(x+1)= (x+1)-1= x and f(f^-1)(x))= f(x-1)= (x-1)+ 1= x.

Or do you simply conclude this qualitatively? I'm thinking if it's just qualitative, you could say something like this:

If (x + 1) - 1 = x, then it must be that + 1 and - 1 do opposite things to x. Thus, you can conclude that (x + 1) and (x - 1) are inverse functions, seeing as they do opposite things to x.

Yes, that's a very good, less technical, way of thinking about inverse functions: whatever f "does", f^-1 "undoes".

Does that sound right? Is there I way to mathematically show this? I know that if you take (x + 1) and (x - 1) and sub one into the other you get either (x + 1) - 1 = x or (x - 1) + 1 = x, but can you do this backwards, by starting from (x + 1) - 1 = x or (x - 1) + 1 = x?

Thanks in advance for any help I get.

Well, once again, f(f^-1(x))= x and f^-1(f(x))= x are the ways to "mathematically show this". Since applying first one and then the other, whatever one "does" to x, the other "undoes".

You can see why inverse functions are so important: in order to solve the equation f(x)= b, you must "back out" and "undo" whatever f has "done" to x. If you know the inverse function of f, you only need to apply that function to both sides of the equation: f^-1(f(x))= x= f^-1(b).

Of course, solving equations, in general, isn't that easy! That's because most functions don't have inverses. If f(x₁) and f(x₂) both equal y, what is f^-1(y)? Is it x₁ or x₂? Since there is no way to answer that, only "one to one" functions have inverse. Further, if there is no x such that f(x)= y, then there can be no f^-1(y). Only "one to one" and "onto" functions have inverses. Since (2)²= 4 and (-2)²= 4, as simple a function as f(x)= x² does not have an inverse. f(x)= x² is neither "one to one" nor "onto". Instead we "modify" the function to : g(x)= x² if x\ge 0 and is undefined for x< 0. We can say that f is "one to one" and "onto" the set of non-negative numbers. Now, if y is a non-negative number, g^{-1}(y)= \sqrt{y} is defined. Of course, if we have the equation f(x)= x²= -4, precisely because f(x)= x² is not "onto" the set of all real numbers, we have to say there is no solution because f^-1(-4) is not defined. Also, because f(x)= x² is not "one to one", to solve f(x)= x²= 4, we have to re-interpret the equation as g(x)= 4 to get x= 2, then recognize that there are two solutions, 2 and -2.

 

[Byrgg]

Err, HallsofIvy said no to my first question, while everyone else says yes it seems.

log_b (n) = x if and only if b ^ x = n, where b > 0, and b is not equal to one.

How does this define logarithm as the inverse of the exponential? log_b (n) = x and b ^ x = n are the same thing...

Also, is my mathematical explanation here right?(see below)

(x + 1) - 1 = x

If you take (x + 1) as the first function of x, you must apply the inverse to it to obtain x, which is x - 1. So if you sub this into x - 1, you get
(x + 1) - 1 = x, since x - 1 does the opposite thing to x, you just proved that (x - 1) and (x + 1) are inverses.

Does that sound right?

 

[Gagle The Terrible]

Quote : "log_b (n) = x and b ^ x = n are the same thing..."

They are not the same thing, but they are linked by an equivalence relation.
It means that if one is true, the other is and vice versa.

HallsofIvy is right in the sense that logarithm's properties and algebra can be considered separately, without mention of the exponential function.

But, to rigorusly proove certain properties of to just define the function itself, it is a possibility to use the inverse of that particular function if the properties of the inverse funtion are well defined.

And I think your last explanation sounds right, just use HallsofIvy's notation/definition to "write it properly".

Also, his statement gives a damn good explanation of what to be aware of in these situations so listen to the man... :-)

 

[HallsofIvy]

No, I did not say "no" to your first question! What I said was

No, "showing how you write logarithms" wouldn't have anything to do with exponentials, would it?

That was in response to your statement

I think it just shows how you write logarithms.

 

[d_leet]

Also, is my mathematical explanation here right?(see below)

(x + 1) - 1 = x

If you take (x + 1) as the first function of x, you must apply the inverse to it to obtain x, which is x - 1. So if you sub this into x - 1, you get
(x + 1) - 1 = x, since x - 1 does the opposite thing to x, you just proved that (x - 1) and (x + 1) are inverses.

Does that sound right?

If I take f(x) = x + 1
and g(x) = x - 1

then what you have shown is that g(f(x)) = (x+1)-1 = x
However you have NOT shown that these two functions are inverses of each other what you have shown is that g is a left inverse for f, or similarly that f is a right inverse for g.

To show that these two functions are inverses you need to show that
f(g(x)) = g(f(x)) = x

 

[Byrgg]

Aren't log_b (n) = x if and b ^ x = n two different ways of saying the same thing? If you rewrite log_b (n) = x you get b ^ x = n...

 

[Byrgg]

Also, what is the process to work backwards from (x + 1) - x = x to showing that (x + 1) and (x - 1) are inverses? Or has this already been shown?

 

[d_leet]

Also, what is the process to work backwards from (x + 1) - x = x to showing that (x + 1) and (x - 1) are inverses? Or has this already been shown?

I'm going to assume you mean (x + 1) - 1 = x not (x + 1 ) - x = x.

let's say we have the function
f(x) = x + 1
for simplicity let's also say y = f(x)
then y = x + 1.


Now if we want to find the inverse function g(x) = f^-1(x)
Then g(y) = x because I have defined g to be the inverse function of f.

The typical way to find an inverse function is to start with an initial function like
y = f(x)

Now if you can solve for x in terms of y then you have a function in terms of y such as x = g(y) and this function will be the inverse of the original function f, but this is not always possible to do, and actually is very rarely possible to do especially in cases involing several trig functions, and higher order polynomials.

I'll go through an example with y = x + 1

Now for sake of ending up with a function in relatively the same form I am going to interchange x and y so that I will end up with a function for y in terms of x, however I am going to denote y by y' to denote that this is not the same y.

So now I have x = y' + 1
or y' + 1 = x and subtracting 1 from each side leaves
y' = x - 1 so I now have found the function that is the inverse of the original, and you can easily show that if I let y = f(x) and y' = g(x) that f(g(x)) = g(f(x)) = x to more rigorously show that these functions are indeed inverses.

 

[HallsofIvy]

Aren't log_b (n) = x if and b ^ x = n two different ways of saying the same thing? If you rewrite log_b (n) = x you get b ^ x = n...

That true because b^{log_b(x)}= x, if log_b(n)= x, you get b^x= n. In other words, log_b(n)= x is equivalent to b^x= n precisely because log_b(x) and b^x are inverse functions.

 

[Byrgg]

I'm going to assume you mean (x + 1) - 1 = x not (x + 1 ) - x = x.p

Yeah, I did mean to type (x + 1) - 1 = x, sorry about that.

When you started the work, you started simply with f(x) = (x + 1). I Wanted to know how you go about it right from (x + 1) - 1 = x.

Also, what would change if you didn't switch y and x towards the end there?

As for Hallsof Ivy... I think I'm understanding that better now, those two equations are equivalent because taking the log(base b) of n results in x only if b ^ x = n. This can only be true if the logarithm is the inverse of the exponential, and so therefore this must define them as inverses, right?

 

[d_leet]

Also, what would change if you didn't switch y and x towards the end there?

If i hadn't changed y and x, it really wouldn't have made a difference i would have ended up with teh same inverse function, but it would have been in the form of x as a function of y and i would have had x = g(y) = f^-1(y) which would still have been the inverse function and if I wanted this to be in the same form as y = f(x) I can just change the variables to y = g(x) and it really doesn't matter as long as you are consistent and understand what you are doing when you do this. And I apologize if this explanation doesn't make much sense because I don't think it does but I'm having trouble trying to explain this in a way that makes some logical sense.

When you started the work, you started simply with f(x) = (x + 1). I Wanted to know how you go about it right from (x + 1) - 1 = x.

And I'm not quite sure what you mean by this, I tried to you how to get
y = x - 1 as the inverse function of y = x + 1 in my last post, so I think I am just not understanding the questiong you're asking here.

 

[Byrgg]

When you started to show the work, you said "let's say we have the function f(x) = x + 1".

I wanted to know where that came from, I was wondering how you would start at (x + 1) - 1 = x, and then get to the rest.

Also, how would you work with the resulting function if you didn't switch y and x there? Could you show me how that would work out?

 

[d_leet]

When you started to show the work, you said "let's say we have the function f(x) = x + 1".

I wanted to know where that came from, I was wondering how you would start at (x + 1) - 1 = x, and then get to the rest.

If I start from there I have (x + 1) - 1 = x + 1 - 1 = x + (1 - 1) = x + 0 = x.

All that really says is that -1 is the additive invers of 1.

But I guess you could look at it like this
give (x + 1) - 1 = x

Now let u = x + 1 you could say that u = u(x)

Then we have u - 1 = x

Now let's say that v = u - 1 but it wouldn't really change this equation if I say that u = x, note that this has nothing to do with the previous function I defined, so I now have v = x - 1 = v(x)

and so I have v(x) = x -1 so let's substitute u(x) into this giving
v(u(x)) = u - 1 = u(x) - 1 = (x + 1) - 1 = x saying that v is a left inverse of u ot u is a right inverse of v.

To me this seems incredibly messy and awkward because I started with an equation, a statement that I can't change and built from this functions, I really don't like this and this process doesn't really seem logical to me. It's preferable to find an inverse the way I stated in one of my previous posts.

Also, how would you work with the resulting function if you didn't switch y and x there? Could you show me how that would work out?

Again I don't really understand your question here. I thought that I tried to explain this in my last post.

 

[Byrgg]

Now let's say that v = u - 1 but it wouldn't really change this equation if I say that u = x, note that this has nothing to do with the previous function I defined, so I now have v = x - 1 = v(x)

That part there kind of confused me, more specifically when you said "but it wouldn't really change this equation if I say that u = x, note that this has nothing to do with the previous function I defined".

I didn't really understand that part, could you explain it please?

Also, why would you assign (x + 1) the value 'u' like you did?

Now about the reversal of x and y, if you didn't reverse them, it would've ended up like this, right?

y = x + 1
y - 1 = x

What would you have done with this relationship? That is more what I'm asking for this part.

 

[d_leet]

That part there kind of confused me, more specifically when you said "but it wouldn't really change this equation if I say that u = x, note that this has nothing to do with the previous function I defined".

What was the u = x part about?

Also, why would you assign (x + 1) the value u like you did?

Because this thread is about inverse functions and the only way I was able to get inverse functions from the equation you wanted to start with was to assign some part of that function a new name. I also said in my last post that I really didnt like this way of finding two inverse functions, it was messy and I'm sure that not everything in that post logically followed from what came before it.

Now about the reversal of x and y, if you didn't reverse them, it would've ended up like this, right?

y = x + 1
y - 1 = x

What would you have done with this relationship? That is more what I'm asking for this part.

Yes so you have that x = y - 1 this is x as a function of y and goes back to what I said earlier that if y = f(x) then x = g(y) where g is the inverse of f. However, if I wanted them both to be functions of x and to maybe graph this so that it looks different than the original function( specifically it is a reflection about the line y = x which is probably better justification for interchanging x and y to find the inverse function) you should change the places of x and y. Again this is a rather poor explanation on my part, but it makes some sense, and hopefully enough to try and explain the problem.

 

[Byrgg]

Because this thread is about inverse functions and the only way I was able to get inverse functions from the equation you wanted to start with was to assign some part of that function a new name. I also said in my last post that I really didnt like this way of finding two inverse functions, it was messy and I'm sure that not everything in that post logically followed from what came before it.

Ok, that answered the last part of that quote, which was: "Also, why would you assign (x + 1) the value 'u' like you did?", but what about the first part of the question I asked in that quote?

Yes so you have that x = y - 1 this is x as a function of y and goes back to what I said earlier that if y = f(x) then x = g(y) where g is the inverse of f. However, if I wanted them both to be functions of x and to maybe graph this so that it looks different than the original function( specifically it is a reflection about the line y = x which is probably better justification for interchanging x and y to find the inverse function) you should change the places of x and y. Again this is a rather poor explanation on my part, but it makes some sense, and hopefully enough to try and explain the problem

So then if you didn't switch x and y, from f(x) = x and g(y) = x, you would get:

f(y - 1) = x and g(x + 1) = x

What would you do from there?

Maybe this?

Since f(y - 1) = x and g(x + 1) = x

f(y - 1) = g(x + 1)

And then what? I don't think there's much else to do...

 

[d_leet]

So then if you didn't switch x and y, from f(x) = x and g(y) = x, you would get:

f(y - 1) = x and g(x + 1) = x

What would you do from there?

Maybe this?

Since f(y - 1) = x and g(x + 1) = x

f(y - 1) = g(x + 1)

And then what? I don't think there's much else to do...

I'll assume you meant f(x) = y, but none of what you have here makes any sense at all, I don't see how you get to f(y - 1) = x and g(x + 1) = x, and yes if they both equal the same value then you can equate them by something called the transitive property. Beyond this I have no clue about what you did here.

 

[Byrgg]

I'll assume you meant f(x) = y, but none of what you have here makes any sense at all, I don't see how you get to f(y - 1) = x and g(x + 1) = x, and yes if they both equal the same value then you can equate them by something called the transitive property. Beyond this I have no clue about what you did here.

Ok, woops, f(x) = y, but then what value can be subbed into x? Didn't you do something like y = x + 1, let y = f(x), so by switching it around, you get x = y - 1. So then f(y - 1) = y, right?

Also, g(y) = x, so then what value can be subbed in for y here? Since y = x + 1, couldn't you sub in the value for y to get g(x + 1) = x?

Oh, and there's still part of one of my questions which remains unanswered(see beginning of last post).

 

[d_leet]

Ok, woops, f(x) = y, but then what value can be subbed into x? Didn't you do something like y = x + 1, let y = f(x), so by switching it around, you get x = y - 1. So then f(y - 1) = y, right?

Also, g(y) = x, so then what value can be subbed in for y here? Since y = x + 1, couldn't you sub in the value for y to get g(x + 1) = x?

Oh, and there's still part of one of my questions which remains unanswered(see beginning of last post).

First as to y = f(x) this is NOTATION, that means y is a function of x, I'll give an example

say f(x) = 3x² + 17x³

I can substitute anything into f(x), like 1 and end up with a numerical answer so in this case f(1) = 20, you could also plug in u and get
f(u) = 3u² + 17u³, I could even plug in "cat" and get something similar.

Your logic and math in this post is still very messy and rather illogical, but I think your problem is confusing notation.


This should take care of your other question as well, but all I did was to substitute u into make it a bit cleaner despite the fact that I will say this again that that is a really messy way of showing functions are inverses.

 

[HallsofIvy]

Yeah, I did mean to type (x + 1) - 1 = x, sorry about that.

When you started the work, you started simply with f(x) = (x + 1). I Wanted to know how you go about it right from (x + 1) - 1 = x.

Also, what would change if you didn't switch y and x towards the end there?

At some point, you must "switch y and x". I personally prefer to switch at the beginning. (So I won't forget at the end!) That's the whole point of "inverse function". If f is some function (such as x+ 1) the writing y= f(x) means: "if we apply the function f to the number x, the result is the number y". Specifically, if y= x+1 and x= 3 then y= 3+1= 4.
That is, the function changes the number 3 to the number 4. The inverse function must do the opposite; change the number 4 to the number 3.
Another way of look at it is to say that if we were graphing this function, we would plot points (x,y)= (x,f(x)) where the second number, y, is the number we get by applying f to the first number, x. The graph of y= f(x)= x+1 must contain the point (3,4) and so the graph of its inverse function, y= f^-1(x) must contain the point (4,3). (The graph of y= f^-1(x) has the nice geometric property that it is the reflection of the graph of y= f(x) around the line y= x.)

In order to find the inverse of y= f(x), a very basic method is: first switch x and y, x= f(y), and then solve for x. If f(x)= x+ 1, switch x and y to get x= y+ 1 and solve for y- subtracting 1 from both sides x-1= y so y= f^-1(x)= x-1. N

"Switch x and y" is not one of the things you normally do while solving equations because the whole point here is not to get an "equivalent" equation, one satisfied by the same values, involving the same function, but rather to get an equation involving a new function, the inverse.

Another example: if f(x)= 3x-2 then y= 3x-2. Switch x and y to get
x= 3y- 2. Now solve for y: x+ 2= 3y so y= (x+2)/3. f^-1(x)= (x+2)/3.

We've already pointed out that the inverse function to f(x)= x+ 1 is f^-1(x)= x- 1 because f is just the "add one" function and "subtract one" is the opposite of that. This new example is just slightly harder than that. f(x)= 3x-2 says "first multiply x by 3 then subtract 2". The opposite of "multiply by 3" is "divide by 3" and the opposite of "subtract 2" is "add 2". To "undo" f, I have to do the opposite operations in the opposite order: first add 2 then divide by 3.

Finding the inverse function is exactly the same as solving the equation y= f(x). Of course, some equations, such as y= x², can have more than one solution so have no inverse function.

As for Hallsof Ivy... I think I'm understanding that better now, those two equations are equivalent because taking the log(base b) of n results in x only if b ^ x = n. This can only be true if the logarithm is the inverse of the exponential, and so therefore this must define them as inverses, right?

 

[Byrgg]

I think I'm understanding most of this pretty well now but here's what I'm really wondering:

When you look at (x + 1) - 1 = x, d_leet assigned the variable u to (x + 1), I'm curious as to why you can do that in the first place. In an example like this, it may be easy to simply assign part of it a value like this, but what if the equation was much more complex than this? I don't know if you can simply assign part of it a value.

 

[HallsofIvy]

I think I'm understanding most of this pretty well now but here's what I'm really wondering:

When you look at (x + 1) - 1 = x, d_leet assigned the variable u to (x + 1), I'm curious as to why you can do that in the first place. In an example like this, it may be easy to simply assign part of it a value like this, but what if the equation was much more complex than this? I don't know if you can simply assign part of it a value.

Why you can do it or why you want to do it? Obviously you can assign a single variable to anything you want. d_leet did that because the whole question was about the inverse of the function f(x)= x+ 1 and he wanted to single that out.
You could find the inverse of f(x)= x+1 by just solving for x:
Subtracting 1 from both sides, x= f(x)- 1. Remembering that the inverse function must satisfy f(f(x))= x, it is clear that the inverse function is "subtract 1". In "function notation", f^-1(x)= x-1 where I have just replaced f(x) with x to get the standard notation.

Because it is simpler to manipulate a single letter than something like "f(x)", d_leet used u= x+1 instead of f(x)= x+1. Solving u= x+ 1 for x is easy: x= u-1 so f^-1(u)= u-1 or, more standard notation, f^-1(x)= x-1.

Because the most common notation is y= f(x), I personally would have used y instead of u: y= f(x)= x+1. Now, (again a personal choice) go ahead and switch x and y: x= y+ 1 (this is the step that changes the function to the inverse function). Solving for y, y= x- 1= f^-1(x).

I don't know if you have read my last post, but I will repeat the part that is relevant to "what if the equation was much more complex than this? "

Suppose we want to find the inverse function to f(x)= 3x+ 2. Write this as y= 3x+ 2. (You say "I don't know if you can simply assign part of it a value." Yes, of course, you can. You can assign a variable to any part of anything!) Switch x and y to get x= 3y+ 2. Now solve for y: 3y= x- 2 so y= (x-2)/3. The inverse function is f^-1(x)= (x-2)/3.

You could write 3((x-2)/3)+ 2= x or ((3x+2)-2)/3= x but you don't have to see that initially in order to find the inverse function.

An even more complicated example: f(x)= 5(2^{3x+2})- 7. Write this as y= 5(2^{3x+2})- 7 and solve for x:
y+ 7= 5(2^{3x+2})
\frac{y+7}{5}= 2^{3x+2}
log_2\left(\frac{y+7}{5}\right)= 3x+2
because log_2(x) is the inverse of 2^x.
log_2\left(\frac{y+7}{5}\right)- 2= 3x
\frac{log_2\left(\frac{y+7}{5}\right)- 2}{3}= x
Finally, since we didn't switch x and y to begin with, we need to do that in order to be able to write this in the "standard form":
f^{-1}(x)= \frac{log_2\left(\frac{x+7}{5}\right)- 2}{3}

 

[Byrgg]

Why you can do it or why you want to do it? Obviously you can assign a single variable to anything you want. d_leet did that because the whole question was about the inverse of the function f(x)= x+ 1 and he wanted to single that out.

But say you didn't know what to single out, say you didn't know what the inverses were for (x + 1) - 1 = x, what would you do then? It's easy to single out a part of this equation, simply because we already know the inverses involved, but in a situation where you don't know the inverses, how would you determine what to single out?

 

[HallsofIvy]

But say you didn't know what to single out, say you didn't know what the inverses were for (x + 1) - 1 = x, what would you do then? It's easy to single out a part of this equation, simply because we already know the inverses involved, but in a situation where you don't know the inverses, how would you determine what to single out?

"What the inverses were for (x+1)- 1= x" makes no sense. Functions have inverses. What you "single out" depends on what function you are talking about. You seem to think that you start with something like "(x+1)- x= x" and "single out" parts that make an inverse. You start with something like f(x)= x+ 1 and then show, as I have explained above, that f^-1(x)= x-1. After you know the inverse, then you know that f^-1(f(x))= x or, in this case, (x+1)-1= x. You also know that f(f^-1(x))= x or, in this case, (x-1)+ 1= x.

 

[Byrgg]

But I want to know what you do if you don't know what function you're dealing with, is there now way to take (x + 1) - 1 = x and figure out the inverses from there?

 

[Hootenanny]

(x + 1) - 1 = x

You do realize that you have just written x = x right?

 

[d_leet]

But I want to know what you do if you don't know what function you're dealing with, is there now way to take (x + 1) - 1 = x and figure out the inverses from there?

What do you mean you don't know what function you're dealing with? HallsofIvy clearly explained that you don't start with an equation to find inverse functions you start with a function and follow the process that he and I both outlined in previous posts. Equations do not have inverses, functions do, and also didn't I already attempt to do this for you in one of my previous posts and I know that I also pointed out that this is really not the way to find inverse functions when I did it.

 

[Byrgg]

So then if you are given and equation in the form f(g(x)) = x, you cannot determine the functions f(x) and g(x)?

Basically, I know how to find the inverse of a function(reversing x and y, and then rearranging to find y), and from this you can obtain f(g(x)) = x. What I find strange though, is why you can't do the opposite, and take f(g(x)) = x to find the functions.

 

[daveb]

Because there are an infinite number of functions that satisfy having an inverse. Suppose you're given exactly what you stated, that is, suppose the only ifo you have is a function f(x) has an inverse so that f(g(x))=g(f(x))=x. Then f(x) = ax, and g(x) = x/a (assume a not equal 0) are inverses of each other and could be your two functions. But so could f(x) = x + b and g(x) = x - b. On the other hand, (again, assume a not equal 0), you could have f(x) = ax + b, and g(x) = (x-b)/a. There are many more funtions which have inverses as well.

 

[HallsofIvy]

So then if you are given and equation in the form f(g(x)) = x, you cannot determine the functions f(x) and g(x)?

Basically, I know how to find the inverse of a function(reversing x and y, and then rearranging to find y), and from this you can obtain f(g(x)) = x. What I find strange though, is why you can't do the opposite, and take f(g(x)) = x to find the functions.

Because there are many different ways to "dissassemble" an equation.
If the equation were the very slightly more complicated (3(x+ 1)- 3)/3= x,
then I could think of that as f(x)= 3(x+1) and f^-1(x)= (x-3)/3.
Then f^-1(f(x))= (f(x)- 3)/3= (3(x+1)-3)/3= x and f(f^-1(x))= 3(f^-1(x))+ 1)= 3((x-3)/3+ 1)= 3(x/3)= x.

But I could, just as well, take f(x)= 3(x+ 1)- 3 and f^-1(x)= x/3. Once again, f^-1(f(x))= f(x)/3= (3(x+1)-3)/3= (x+1)- 1= x and f(f^-1(x))= 3(f^-1(x)+ 1)- 3= 3(x/3- 1)- 3= (x+ 3)- 3= x.

If I told you that x= 12 and y= 9, you could tell me that xy= 108.
If I told you that xy= 108, could you tell me what x and y are?

 

[Byrgg]

Here's something I'm wondering, you can prove the equation (x + 1) - 1 = x simply, 1 - 1 = 0, 0 + x = x, easy. But say you have something like b^{log_b (x)} = x, how can you simplify this as I did with the previous equation?

 

[HallsofIvy]

Here's something I'm wondering, you can prove the equation (x + 1) - 1 = x simply, 1 - 1 = 0, 0 + x = x, easy. But say you have something like b^{log_b (x)} = x, how can you simplify this as I did with the previous equation?

What do you mean by "simplify"? it seems pretty obvious that you could write f(x)= b^x and f^-1= log_b(x) but that's just using the definition of b^x and log_b(x). It's also true that b(b^{log_b(x)-1})= x so you could also write f(x)= b^{x-1}
and f^{-1}(x)= b log_b(x). You have to have a function in order to have an "inverse" function, not an equation.

 

[Byrgg]

What do you mean by "simplify"?

I meant simplify as I did in my example:

(x + 1) - 1 = x
x + 1 - 1 = x
x + (1 - 1) = x
x + 0 = x
x = x

I'm wondering if there's a similar way to do this with b^{log_b (x)} = x.

It's also true that b(b^{log_b(x)-1})= x so you could also write f(x)= b^{x-1}
and f^{-1}(x)= b log_b(x). You have to have a function in order to have an "inverse" function, not an equation.

Was there a typo or something in this example? I didn't really understand it, maybe I'm just not reading it right or something.

 

[Hootenanny]

I meant simplify as I did in my example:

(x + 1) - 1 = x
x + 1 - 1 = x
x + (1 - 1) = x
x + 0 = x
x = x

I'm wondering if there's a similar way to do this with b^{log_b (x)} = x.



Was there a typo or something in this example? I didn't really understand it, maybe I'm just not reading it right or something.

Again, both of those (x + 1) - 1 = x and b^{log_b (x)} = x are equations. EQUATIONS DO NOT HAVE INVERSES only functions have inverses, this has been said many times before ensure that you know the difference between a function and an equation.

 

[HallsofIvy]

I meant simplify as I did in my example:

(x + 1) - 1 = x
x + 1 - 1 = x
x + (1 - 1) = x
x + 0 = x
x = x

I'm wondering if there's a similar way to do this with b^{log_b (x)} = x.

Only by noting that by definition b^{log_b (x)} = x, just as by definition 1-1= 0!

Was there a typo or something in this example? I didn't really understand it, maybe I'm just not reading it right or something.

No, there was no typo. My point was that yes, you could "disassemble" the equation b^{log_b (x)} = x and choose to make f(x)= b^x and f^-1(x)= log_b(x). But you could also choose to write b^{log_b(x)}as b (b^{log_b(x)-1}, taking out 1 &amp;quot;b&amp;quot; separately. Then you could assert that by taking f(x)= b(b^x) and f&lt;sup&gt;-1&lt;/sup&gt;(x)= log&lt;sub&gt;b&lt;/sub&gt;(x)-1, you still have f(f&lt;sup&gt;-1&lt;/sup&gt;(x))= x.&lt;br /&gt; &lt;br /&gt; One more time: Given any equation of the form F(x)= x, there is &lt;b&gt;not&lt;/b&gt; a unique way to &amp;quot;disassemble&amp;quot; F(x) into f(x) and f&lt;sup&gt;-1&lt;/sup&gt;(x). You must &lt;b&gt;start&lt;/b&gt; with a function f(x) and seek its inverse function, f&lt;sup&gt;-1&lt;/sup&gt;(x), if any.

 

[Byrgg]

But you could also choose to write b^{log_b(x)}as b (b^{log_b(x)-1}, taking out 1 &quot;b&quot; separately. Then you could assert that by taking f(x)= b(b^x) and f-1(x)= logb(x)-1, you still have f(f-1(x))= x.

<br /> <br /> I&#039;m not really understanding this part very well, could you clarify it a bit?

 

[benorin]

Should read:

b^{log_b(x)} as b (b^{log_b(x)-1})

 

[HallsofIvy]

Byrgg, since you don't seem to understand what we have said repeatedly, I'm going to get even simpler- numbers instead of functions. Please don't be offended if it seems trivial, it addresses exactly the problem you seem to be having.

You know that the "additive inverse" of 5 is -5. That's true because 5+(-5)= 0 and (-5)+ 5= 0. Now, suppose we are given the equation 5+ (-5)= 0. Can we immediately conclude that -5 is the additive inverse of 5? Well, yes, I guess that it does tell us that!

But suppose instead, we have 4+ 3- 7= 0. What can we conclude, about additive inverses from that? I suppose writing 4+ 3 as 7 would be obvious: 7+ (-7)= 0 so that equation tells us that the additive inverse of 7 is -7 (obviously a true statement). But I could just as easily write 3- 7 as -4 and interpret "4+ 3- 7= 0" as "4+ (-4)= 0" and conclude that the equation tells us that "the additive inverse of 4 is -4" (also a true statement. It takes a little more arithmetic but I could write 5 as 9- 4 and -5 as 2- 7 so that 5- 5 becomes (9- 4)+ (2-7) leading to the conclusion that (9+ 2)+ (-4-7)= 11+ (-11)= 0. Aha! the equation tells me that the additive inverse of 11 is -11! In other words, the equation 5+ (-5)= 0 can manipulated to give a+ (-a)= 0 for any numbers a.

I have repeatedly said that there is no one unique way to "disassemble" an equation. Given two numbers, say 5 and 3, there is only one way to add them: 5+ 3= 8. But if I am given the number 8, there are an infinite number of ways to "disassemble" it- to find two numbers, x and y, such that x+ y= 8: 4+ 4= 8, -10+ 18= 8, etc.

Similarly, given two functions, f and g, there is only one way to compose them: f(g(x))= h(x) is unique. But given a function h(x) (even if it is simply h(x)= x) there are an infinite number of ways to find f and g so that f(g(x))= x.

Given an equation like (x+ 1)- 1= x, it is obvious to say f(x)= x+ 1, f^-1(x)= x-1 but we could just as easily manipulate it into other things: adding 2 and -2, (x+ 1)+ 2- 2- 1= x becomes (x+ 3)- 3= x so f(x)= x+ 3, f^-1(x)= x- 3.

One last time: given h(x)= x, there are an infinite number of ways to algebraicly manipulate h into "f(g(x))" so that we can conclude that f and g are inverse functions.

 

[Byrgg]

I know that functions are not equations, maybe I wasn't clear on what I was asking. HallsofIvy has said that there is no unique way to determine the functions involved when given an equation, given the fact that there are infinite ways to disassemble an equation. What has confused me in all of this, is that in my first thread, I believe it HallsofIvy who said b^{log_b x} = x precisely said that b^x and log_b(x) were inverses. After hearing the explanations here, it seems that pulling these functions out like that was wrong, since there were infinite ways to disassemble the equation. Could someone clear this up please?

It takes a little more arithmetic but I could write 5 as 9- 4 and -5 as 2- 7 so that 5- 5 becomes (9- 4)+ (2-7) leading to the conclusion that (9+ 2)+ (-4-7)= 11+ (-11)= 0. Aha! the equation tells me that the additive inverse of 11 is -11! In other words, the equation 5+ (-5)= 0 can manipulated to give a+ (-a)= 0 for any numbers a.

Is this still part of the example you were working with, or did you start a new example? If it's still part of the last example, then I'm lost.

But you could also choose to write b^{log_b(x)}as b (b^{log_b(x)-1}, taking out 1 &quot;b&quot; separately. Then you could assert that by taking f(x)= b(b^x) and f-1(x)= logb(x)-1, you still have f(f-1(x))= x.

<br /> <br /> I&#039;m still not understanding this part very well, could you clarify it a bit? I&#039;m not really sure what is being done here.

 

[Office_Shredder]

I know that functions are not equations, maybe I wasn't clear on what I was asking. HallsofIvy has said that there is no unique way to determine the functions involved when given an equation, given the fact that there are infinite ways to disassemble an equation. What has confused me in all of this, is that in my first thread, I believe it HallsofIvy who said b^{log_b x} = x precisely said that b^x and log_b(x) were inverses. After hearing the explanations here, it seems that pulling these functions out like that was wrong, since there were infinite ways to disassemble the equation. Could someone clear this up please?

There are infinite ways to dsassemble the equation, we choose to use the most convenient way. Would you rather do 5 + (-5) by doing 5-5, or by doing (5+5)/2 + (1-16)/3?

They're both accurate ways to break the equation down, it's just that one is harder

Is this still part of the example you were working with, or did you start a new example? If it's still part of the last example, then I'm lost.

It's part of the last example, showing how many ways there are to break up an equation

I'm still not understanding this part very well, could you clarify it a bit? I'm not really sure what is being done here.

Ignore it, it's not necessary

 

[Byrgg]

It's part of the last example, showing how many ways there are to break up an equation

I still don't really understand how that part came about, could someone explain please?

Ignore it, it's not necessary

I don't really want to ignore it, could someone clarify it please?

 

[Office_Shredder]

Since b^y = b*b^y-1, the last part comes trivially by replacing y with log_bx

 

[HallsofIvy]

You had started from the equation b^{log_b(x)}= x and immediately got that f(x)= b^x and f^-1(x)= log<sub>b[/sup](x) are inverse functions. My point was that an equation like that involves one function, not two. There is no unique way to break a single function into two.

We could, almost as easily, write b^{log_b(x)}= x as b^{log_b(x)+ 1- 1}= b^1 b^{log_b(x)-1}= x and think of that as f(x)= b b^x and f^-1(x)= logb(x)- 1.</sub>

 

[Byrgg]

b^{log_b(x)}= x[/itex]and immediately got that f(x)= bx and f-1(x)= log_{b[/sup](x) are inverse functions. My point was that an equation like that involves one function, not two. There is no unique way to break a single function into two.}

_{I thought that b^{log_b(x)}= x involved two functions, considering that it's in the form g(f(x)) = x, isn't it? And I believe it was you who originally got b^x and log_b(x) are inverse functions from b^{log_b(x)}= x, as stated in a post in an earlier thread. Sorry if I misinterpreted or anything.}