# Doubling Time from population growth equation

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• Sep 29th 2007, 10:30 AM
conor
Doubling Time from population growth equation
I need to isolate the variable n (years) from this equation:

Pt = Po * (1 + {GR/100})^n

I tried using logs:

log Pt = log Po * n log (1 + {GR/100})

then I tried n = _________log Pt_________
log Po * log (1 + {GR/100})

Dows anyone know where I am going wrong? Its been ages since i've used logs.
• Sep 29th 2007, 01:21 PM
ticbol
Your solution looks good to me. If the expressions were to be posted aligned, it could be better.

You meant
n = Log[Pt] / {Log[Po] *Log[1 +(GR /100)]}
right?

You're still good after all those ages.
Because the Log[1 +(GR/100)] cannot be touched anymore.
• Sep 29th 2007, 01:35 PM
conor
awesome! but can it be furthur simplified, or is that the end of the equation... at least its usable.
• Sep 29th 2007, 01:54 PM
ticbol
I cannot simplify it any further.

Log(a) *Log(b) is just that.
It cannot be Log(a*b), no.
Nor Log(a +b), no.

Log(a +b) is just that also.
It cannot be Log(a) +Log(b), no.
Nor Log(a) *Log(b), no.
• Sep 29th 2007, 02:07 PM
conor
Is this right, then?

http://img9.imagevenue.com/loc623/th..._122_623lo.jpg

Thanks for all your help!
• Sep 29th 2007, 03:00 PM
ticbol
Quote:

Originally Posted by conor
Is this right, then?

http://img9.imagevenue.com/loc623/th..._122_623lo.jpg

Thanks for all your help!

Whoa.

Now that I've seen it like that, I'm sorry to mislead you.

No, that is wrong a little bit.

It should be like this:

Pt = Po*[1 +(GR/100)]^n

Log(Pt) = Log(Po) +n*Log[1 +(GR/100)]
Log(Pt) -Log(Po) = n*Log[1 +(GR/100)]
Log[(Pt) /(Po)] = n* Log[1 +(GR/100)]

n = Log[Pt / Po] / Log[1 +(GR/100)] ---------------answer.

--------------------------------------------

Again, sorry. I was fresh from sleeping when I commented before.
• Sep 30th 2007, 12:29 PM
conor
ohhhh because a = b*c
then log[a] = log[b] + log[c]

is that right?
• Sep 30th 2007, 12:42 PM
conor