solving for t in compound interest equation

[jonboy]

the question is: solving for t in compound interest equation

this is not an assignment for school, but a question that i wondered upon myself.

I though of a way, but I'm not too sure. Here's my shot:

We have: \(\displaystyle A(t)\,=\,P(1\,+\,\frac{r}{n})^{nt}\)

First divide by P:\(\displaystyle \frac{A(t)}{P}\,=\,(1+\frac{r}{n})^{nt}\)

Take the n root of both sides: \(\displaystyle \sqrt[n]{{\frac{{A(t)}}{P}}} = \left( {1 + \frac{r}{n}} \right)^t\)

Change to logarithmic form: \(\displaystyle log_{(1+\frac{r}{n})}\sqrt[n]{\frac{A(t)}{P}}\,=\,t\)

Is this correct? Taking the n root of both sides is what I'm iffy on.

Later on I'm going to explore on solving for n. That'll be tough as it is on the right sides of the equation twice.
I googled this and I did not get any help.

 

[soroban]

Hello, jonboy!

Solving for \(\displaystyle t\) in compound interest equation

\(\displaystyle \text{We have: }\;P\left(1 +\frac{r}{n}\right)^{nt} \;=\;A\)

\(\displaystyle \text{First divide by }P\!:\;\left(1+\frac{r}{n}\right)^{nt}\;=\;\frac{A}{P}\)

\(\displaystyle \text{Take the }n^{th}\text{ root of both sides: }\;\left( {1 + \frac{r}{n}} \right)^t \;=\;\sqrt[n]{\frac{A}{P}} \;=\;\left(\frac{A}{P}\right)^{\frac{1}{n}}\)

Take the log of both sides.
. . We can use any base . . . I'll use base "e".

\(\displaystyle \ln\left(1 + \frac{r}{n}\right)^t \;=\;\ln\left(\frac{A}{P}\right)^{\frac{1}{n}} \quad\Rightarrow\quad t\cdot\ln\left(1 + \frac{r}{n}\right) \;=\;\frac{1}{n}\cdot\ln\left(\frac{A}{P}\right)\)

. . \(\displaystyle \text{Therefore: }\;\boxed{t \;=\;\frac{\ln\left(\dfrac{A}{P}\right)}{n\cdot\ln\left(1 + \dfrac{r}{n}\right)}}\)

 

[jonboy]

thanks Soroban! it's always nice to see you.

i'll ponder this later today when my brain's fired up.