Solve for t using logarithms?

**iheartthemusic29** · September 4th 2009, 07:03 PM

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

**Rapha** · September 4th 2009, 08:42 PM

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

$Wf^t=Qb^t$

$\frac{f^t}{b^t} = \frac{Q}{W}$

$\displaystyle (\frac{f}{b})^t = \frac{Q}{W}$

$LN((\frac{f}{b})^t) = LN(\frac{Q}{W})$

$t*LN(\frac{f}{b}) = LN(\frac{Q}{W})$

$t= \frac{LN(\frac{Q}{W})}{LN(\frac{f}{b})}$

$t= \frac{LN(\frac{Q}{W})}{LN(f)-LN(b)}$

$t= \frac{LN(Q)-LN(W)}{LN(f)-LN(b)}$

Any questions?

regards
Rapha

**HallsofIvy** · September 5th 2009, 03:29 AM

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)

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