Getting r or t out of e^rt

**dexteronline** · September 24th 2011, 05:30 AM

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)]

**Plato** · September 24th 2011, 05:44 AM

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

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

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

**dexteronline** · September 24th 2011, 06:18 AM

Originally Posted by Plato

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

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

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

You solved my problem, thank you

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