Wf^t=Qb^t
I solved a similar problem that involved natural logs, but this one confused me more.
Can someone help me?
Hi iheartthemusic29.
It is a very good idea to use logarithms, lets see
$\displaystyle Wf^t=Qb^t$
$\displaystyle \frac{f^t}{b^t} = \frac{Q}{W}$
$\displaystyle \displaystyle (\frac{f}{b})^t = \frac{Q}{W}$
$\displaystyle LN((\frac{f}{b})^t) = LN(\frac{Q}{W})$
$\displaystyle t*LN(\frac{f}{b}) = LN(\frac{Q}{W})$
$\displaystyle t= \frac{LN(\frac{Q}{W})}{LN(\frac{f}{b})}$
$\displaystyle t= \frac{LN(\frac{Q}{W})}{LN(f)-LN(b)}$
$\displaystyle t= \frac{LN(Q)-LN(W)}{LN(f)-LN(b)}$
Any questions?
regards
Rapha
I think I would be inclined to just take the logarithm directly:
log(Wf^t)= log(Qb^t)
log(W)+ log(f^t)= log(Q)+ log(b^t)
log(W)+ t log(f)= log(Q)+ t log(b)
log(W)+ t log(f)- t log)b)= log(Q)
t(log(f)- log(b)= log(Q)- log(W)
t log(f/b)= log(Q/W)
t= log(Q/W)/log(f/b)