# Getting r or t out of e^rt

• Sep 24th 2011, 05:30 AM
dexteronline
Getting r or t out of e^rt
For a function such as (1+r)^n = FV/PV one can take the Log of both sides to find n

nlog (1+r) = log(FV/PV)
nlog (1+r) = log(FV)-log(PV)
n = [log(FV)-log(PV)]/log (1+r)

I need a bit of assistance in solving similar problem that has e instead of (1+r)

e^rt = FV/PV

What has me confuses here is the relationship between log and e as each of these is the inverse function of the other

So do I use the same logic as I did with the earlier stated problem

rt log(e) = log(FV/PV)
rt log(e) = log(FV)-log(PV)
t = [log(FV)-log(PV)]/[r log(e)]
• Sep 24th 2011, 05:44 AM
Plato
Re: Getting r or t out of e^rt
Quote:

Originally Posted by dexteronline
For a function such as (1+r)^n = FV/PV one can take the Log of both sides to find n
nlog (1+r) = log(FV/PV)
nlog (1+r) = log(FV)-log(PV)
n = [log(FV)-log(PV)]/log (1+r)
I need a bit of assistance in solving similar problem that has e instead of (1+r)
e^rt = FV/PV
rt log(e) = log(FV/PV)
rt log(e) = log(FV)-log(PV)
t = [log(FV)-log(PV)]/[r log(e)]

Historically $\displaystyle \log(x)$ meant $\displaystyle \log_{10}(x)$ and $\displaystyle \ln(x)$ meant $\displaystyle \log_e(x)$

Thus in your problem $\displaystyle rt\ln(e)=rt$ because $\displaystyle \ln(e)=1$.

Here is a warning: many today $\displaystyle \log$ to be $\displaystyle \log_e=\ln(x)$.
• Sep 24th 2011, 06:18 AM
dexteronline
Re: Getting r or t out of e^rt
Quote:

Originally Posted by Plato
Historically $\displaystyle \log(x)$ meant $\displaystyle \log_{10}(x)$ and $\displaystyle \ln(x)$ meant $\displaystyle \log_e(x)$

Thus in your problem $\displaystyle rt\ln(e)=rt$ because $\displaystyle \ln(e)=1$.

Here is a warning: many today $\displaystyle \log$ to be $\displaystyle \log_e=\ln(x)$.

You solved my problem, thank you :)