# Simplifying logarithms

1. Jun 26, 2006

### star321

my question is simplify:
logpie(1-cosx)+logpie(1+cosx)-2logpie sinx (i dont know how to make the pie symbol)

i thought it was going to be
logpie-logpiecosx+logpie+logpiecosx-2logpiesinx
=
2logpie-2logpiesinx

But, from my other post I was told you cannot do that with ln's.. is it the same for logs, that I cannot do that? If not, any suggestions on how i could simplify this?

2. Jun 26, 2006

### arildno

1. Open up your textbook where the rules for the arithmetic of logarithms stand.

2. Post those rules in your next post.

3. THINK about those very rules, and see how you may apply them!

3. Jun 26, 2006

### tim_lou

log(a+b) does not equal log(a)+log(b)
rather,
log(ab)=log(a)+log(b)
be careful

4. Jun 26, 2006

### star321

OK
So laws of logarithms:
1) loga1=0
2)loga(xy)=logax+logay
3) loga (1/x)=-logax
4) loga(x/y)=logax-logay
5)loga(x^y)=ylogax
6)logax=logbx/logba

so
logpie(1-cosx)+logpie(1+cosx)-2logpie sinx

using 2)loga(xy)=logax+logay
logpie(1-cosx)(1+cosx)-2logpiesinx
=-2logpiecosx-2logpiesinx
then from 4) loga(x/y)=logax-logay
logpie=-2cosx/-2sinx
so logpie=cosx/sinx

That seems better to me, or did I make a mistake somewhere?

5. Jun 26, 2006

### arildno

Great, star321!
Now, for your mistakes:
logpie(1-cosx)(1+cosx)-2logpiesinx
=-2logpiecosx-2logpiesinx

This is wrong!
You have:
$$\log_{\pi}((1-\cos(x))(1+\cos(x)))=\log_{\pi}(1-\cos^{2}(x))$$
This is NOT equal to $2\log_{\pi}(\cos(x))$!!

However, we DO have the trigonometric identity $1-\cos^{2}(x)=\sin^{2}(x)$

Therefore, you have:
$$\log_{\pi}((1-\cos(x))(1+\cos(x)))=2\log_{\pi}(\sin(x))$$

Your application of the fraction rule is also faulty.

Last edited: Jun 26, 2006
6. Jun 26, 2006

### tim_lou

just out of curiosity... why did you put pie instead of pi?

7. Jun 26, 2006

### star321

oh is it sposed to be pi?? hehe.. i didnt know.

8. Jun 26, 2006

### TD

Yes, pie is something you eat. You don't eat pi
You could try eating pi pies though

9. Jun 26, 2006

### HallsofIvy

Staff Emeritus
It seems better?? What do you think "$log_\pi$" MEANS??
$log_\pi$ is a function. The function by itself is meaningless:
$log_\pi$ of what?

10. Jun 26, 2006

### 3trQN

Could you? Is it really possible to eat Pi(Pies)?

At some point your pie will need to be divided into such a small fraction that there is no way to seperate the individual pie mixture ( the collection of molecules that can be said to make up a pie ) into that small a part of the whole without it ceasing to be a true pie....

So as you approach the 50th decimal expansion of Pi(Pies) you would be in the realm of protons, and as such, it would be impossible to distinguish a pie from a mellon....

Is this correct?

11. Jun 26, 2006

### arildno

Nor would it be distinguishable from cows' dung.

12. Jun 26, 2006

### tim_lou

if you are in a non-Euclidean space-time where pi=3. You can eat 3 pies!

13. Jun 28, 2006

### 3trQN

How long would it take you to eat them?