Printable View
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.
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.
awesome! but can it be furthur simplified, or is that the end of the equation... at least its usable.
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.
Is this right, then?
http://img9.imagevenue.com/loc623/th..._122_623lo.jpg
Thanks for all your help!
Whoa.Quote:
Originally Posted by conor
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.
ohhhh because a = b*c
then log[a] = log[b] + log[c]
is that right?