# Thread: Doubling Time from population growth equation

1. ## 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.

2. 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.

3. awesome! but can it be furthur simplified, or is that the end of the equation... at least its usable.

4. 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.

5. Is this right, then?

Thanks for all your help!

6. Originally Posted by conor
Is this right, then?

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.

7. ohhhh because a = b*c
then log[a] = log[b] + log[c]

is that right?

8. TADAA!