# Awkard question with logarithms

1. Jun 23, 2006

### Byrgg

$b^(log_b x) = x$

a while back, and recieved 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.

Last edited: Jun 23, 2006
2. Jun 23, 2006

### 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.

3. Jun 23, 2006

### 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).

4. Jun 23, 2006

### 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)

5. Jun 23, 2006

### nrqed

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

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

6. Jun 23, 2006

### 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.

7. Jun 23, 2006

### nrqed

: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

8. Jun 23, 2006

### 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 gotta 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.

9. Jun 23, 2006

### 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?

10. Jun 23, 2006

### 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.

11. Jun 23, 2006

### 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?

12. Jun 23, 2006

### Hootenanny

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

13. Jun 23, 2006

### 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)?

14. Jun 24, 2006

### 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.

15. Jun 24, 2006

### 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?

16. Jun 24, 2006

### Hootenanny

Staff Emeritus
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.

17. Jun 24, 2006

### nrqed

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).

18. Jun 24, 2006

### 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...

19. Jun 24, 2006

### arildno

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

20. Jun 24, 2006

### Hootenanny

Staff Emeritus
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?