Logs are hell!

[Dr_Evil]

I would have considered it but what you wrote wasn't correct

How about Log(10p) then?

Listen to Psiko, he knows...

 

[Biohazard]

How about Log(10p) then?

Tramp lol!

 

[Dave]

Take the log of both sides, simplify (log(xy) = log x + log y, log10=1 etc) and you get your answer. That's your standard exponential formula except you'll come across it as Ae^(kt) in things like radioactivity, so you use natural/Napier logs instead of log to the base 10.

"spoiler"

Taking log of both sides gives: (and calling z(0) = y to simplify things for me!)
log z = log (y*10^(-kt))

Simplifying by log(xy) = log(x) + log(y) gives:
log z = log y + log(10^-kt)

Simplifying using log(x^n) = nlogx gives:
log z = log y + (-kt)log(10)

Now, log(10) = 1 and thus;
log z = log y - kt (as required)

Now, where's my 50p

 

[Tommy B]

It was the LogAB = LogA + logB rule I forgot.

Solution:

Thanks everyone.

 

[Dr_Evil]

Take the log of both sides, simplify (log(xy) = logx + log y, log10=1 etc) and you get your answer. That's your standard exponential formula except you'll come across it as Ae^(kt) in things like radioactivity, so you use natural/Napier logs instead of log to the base 10.

So how do you prove all that is true to a bunch of OcUk'ers?

 

[Al Vallario]

Thanks everyone.

No problem

 

[[ASSE]Hinchy]

Damn! This place needs less comedians and more mathematicians

I'd say it was the other way around to be honest

 

[Dave]

So how do you prove all that is true to a bunch of OcUk'ers?

Write it down, put a loaf of Hovis' finest on it, and take a picture

 

[Tommy B]

Take the log of both sides, simplify (log(xy) = log x + log y, log10=1 etc) and you get your answer. That's your standard exponential formula except you'll come across it as Ae^(kt) in things like radioactivity, so you use natural/Napier logs instead of log to the base 10.

"spoiler"

Taking log of both sides gives: (and calling z(0) = y to simplify things for me!)
log z = log (y*10^(-kt))

Simplifying by log(xy) = log(x) + log(y) gives:
log z = log y + log(10^-kt)

Simplifying using log(x^n) = nlogx gives:
log z = log y + (-kt)log(10)

Now, log(10) = 1 and thus;
log z = log y - kt (as required)

Now, where's my 50p

If I ever meet you, I will buy you a can of coke

Or is it 60p now? I forget.

 

[Dave]

If I ever meet you, I will buy you a can of coke

Or is it 60p now? I forget.

Bar of galaxy will do just as fine

If you're in Sheffield then we can meet up to come to some fiscal arrangement when I get back to university

 

[Biohazard]

Bar of galaxy will do just as fine

If you're in Sheffield then we can meet up to come to some sexual arrangement when I get back to university

**caugh**

 

[Al Vallario]

If you're in Sheffield then we can meet up to come to some fiscal arrangement when I get back to university

He describes his location as "the place to be," so I wouldn't hold your breath

 

[Solari]

Logs are hell, especially when they won't flush down...

 

[Dave]

He describes his location as "the place to be," so I wouldn't hold your breath

. . . touché

 

[Tommy B]

Bar of galaxy will do just as fine

If you're in Sheffield then we can meet up to come to some fiscal arrangement when I get back to university

Really? I may end up there next year. My first two uni choices are Sheffield and Newcastle

 

[Tommy B]

He describes his location as "the place to be," so I wouldn't hold your breath

lol

It's not that bad is it!

It rhymes with my name btw.

 

[gumbald]

I was late into this thread, silly me enjoying the sun all day...

Without knowing the content of the post, I'd have gone for: