# Thread: Solving for rate (removal of logs)

1. ## Solving for rate (removal of logs)

Hi

I am trying to solve for r in the following:

F=P(1+r)^t

I've got it to: log(t) =log(F/P) / Log(1+r)

can someone please show me the step by step stages to make r the subject? How do i remove the logs?

Many thanks

H

2. No need for logarithms...

$\displaystyle F=P(1+r)^t$

$\displaystyle \frac{F}{P} = (1+r)^t$

$\displaystyle \left(\frac{F}{P}\right)^{\frac{1}{t}} = \left[(1+r)^t\right]^{\frac{1}{t}}$

$\displaystyle \left(\frac{F}{P}\right)^{\frac{1}{t}} = 1 + r$

$\displaystyle \left(\frac{F}{P}\right)^{\frac{1}{t}} - 1 = r$.

3. thanks a million for this!

follow up question - how would i solve the equation if i used logs from the stage i mentioned....would it be possible without going through a 10 stage process?

4. You only use logarithms if you are solving for the exponent.

5. Originally Posted by hisab
Hi

I am trying to solve for r in the following:

F=P(1+r)^t

I've got it to: log(t) =log(F/P) / Log(1+r)
This is wrong. log(a^b)= b log(a), not "log(b)log(a)". From
F/P= (1+ r)^t, taking the logarithm of both sides give log(F/P)= t(log(1+ r)) so that log(1+ r)= (1/t) log(F/P)= log((F/P)^{1/t}). Now take the "anti-log" (exponential) of both sides to get 1+ r= (F/P)^{1/t} and so r= (F/P)^{1/t}- 1, exactly what Prove It said.

can someone please show me the step by step stages to make r the subject? How do i remove the logs?

Many thanks

H