# Thread: Solve for t using logarithms?

1. ## Solve for t using logarithms?

Wf^t=Qb^t
I solved a similar problem that involved natural logs, but this one confused me more.
Can someone help me?

2. Hi iheartthemusic29.

Originally Posted by iheartthemusic29
Wf^t=Qb^t
I solved a similar problem that involved natural logs, but this one confused me more.
Can someone help me?
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

3. Originally Posted by iheartthemusic29
Wf^t=Qb^t
I solved a similar problem that involved natural logs, but this one confused me more.
Can someone help me?
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)