Inverse Functions: Logarithms & Qualitative Conclusions

[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 [itex]b^{log_b(x)}[itex] as $b (b^{log_b(x)-1}$, taking out 1 "b" separately. Then you could assert that by taking f(x)= b(b^x) and f^-1(x)= log_b(x)-1, you still have f(f^-1(x))= x.

One more time: Given any equation of the form F(x)= x, there is not a unique way to "disassemble" F(x) into f(x) and f^-1(x). You must start with a function f(x) and seek its inverse function, f^-1(x), if any.

 

[Byrgg]

But you could also choose to write [itex]b^{log_b(x)}$as b (b^{log_b(x)-1}$, taking out 1 "b" 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.

I'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 [itex]b^{log_b(x)}$as b (b^{log_b(x)-1}$, taking out 1 "b" 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.

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.

 

[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.}

 

[HallsofIvy]

One more time and then I'm going to drop this: there is no unique way to divide a formula into two functions. There might be a "simplest" way that seems obvious but that is not a mathematical property.

 

[Byrgg]

One more time and then I'm going to drop this: there is no unique way to divide a formula into two functions. There might be a "simplest" way that seems obvious but that is not a mathematical property.

So then how were you able to say that $b^{log_b(x)}= x$ is saying that b^x and log_b(x) are inverse functions? Later in the thread in which you said this, it seemed this was taken from the definition of the logarithmic function, however, this is what you said early on in that thread:

What you are asked to prove is that $b^{log_b(x)}= x$ , which is precisely saying that bx and logb(x) are inverse functions. HOW you would prove that depends on what definitions of logb(x) and bx you are using.

I am confused by this, since you yourself have said that there is no unique way to divide an equation into two separate functions. However, here it seems as though you have done this. Would you, or someone else be able to clear this up for me?

 

[Office_Shredder]

byrgg, it's like this. Let's say you want to show that 12 is even. The simplest way is to say:

12=6*2, which is even by definition. Or you could do this:

12=4*3
4=2*2 so 12=3*2*2 which is even by definition

While this is a trivialized case, you can see a number of important points here:

1.) There were multiple ways of breaking up the number 12.

2.) One way was three times as fast as the other way in terms of solving the problem

It's the same thing with b^log(x)


For the second part:
Again, the problem makes it clear that you are looking at two functions, b^x and log_bx While you're able to break it up in different ways than that, it's the easiest and simplest way to do it.

Now, the reason why those two functions are best, is that b^x and log_bx are well known inverses, and you're attempting to prove that your function $b^{log_b(x)}= x$
is composed of a function and its inverse (because it equals x). It's composed of an infinite set of functions and their inverses, but b^x and log_bx are the easiest, simplest, and most well known inverses.

And the point of the statement was to ask how you have defined b^x and log_bx, because the definition can vary widely (log can be defined as the inverse of an exponent, for example. This would make your life very simple)

 

[Byrgg]

For the second part:
Again, the problem makes it clear that you are looking at two functions, b^x and log_b(x) While you're able to break it up in different ways than that, it's the easiest and simplest way to do it.

You said that the problem is showing you two functions, whereas HallsofIvy said that the problem consists of only one function. Could you clarify this?

 

[Byrgg]

Someone please respond.