Simplifying Logarithms: Solving logpie(1-cosx)+logpie(1+cosx)-2logpie sinx

[star321]

my question is simplify:
logpie(1-cosx)+logpie(1+cosx)-2logpie sinx (i don't 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?

 

[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!

 

[tim_lou]

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

 

[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?

 

[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.

 

[tim_lou]

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

 

[star321]

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

 

[TD]

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

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

 

[HallsofIvy]

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?

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?

 

[3trQN]

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

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 separate 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...

There are few, if any, cases in engineering and science where more than a few dozen digits are needed; with the 50 digits given here; the circumference of any circle that would fit in the observable universe (ignoring the curvature of space) could be computed with an error less than the size of a proton.

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?

 

[arildno]

Nor would it be distinguishable from cows' dung.

 

[tim_lou]

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

 

[3trQN]

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

How long would it take you to eat them?