My assumption is that you are trying to find out how old the hide is. I'm going with that. The formula is given to us. Our N value is 30, our
![N_(0)](https://img.qammunity.org/2019/formulas/mathematics/college/yjwa9b1nyregxirt4atco6ohd7ye33vpov.png)
is 100, and we are given the constant k as -.0001. Setting up our formula accordingly, we have
![30=100e ^(-.0001t)](https://img.qammunity.org/2019/formulas/mathematics/high-school/129ecwxvbmabaiqncnr0pu7w1ym32bmlmy.png)
. In solving for t, we are left with the problem of getting out from the exponent that it's in to down on a level where we can deal with it. Natural logs and Euler's number "undo" each other, if you will, just to "undo" a square root we would square it. We will take the natural log of both sides to "undo" Euler's number. (It actually is because natural logs have a base of "e", but nonetheless...). First things first, we will divide both sides by 100 to get
![.3=e ^(-.0001t)](https://img.qammunity.org/2019/formulas/mathematics/high-school/3dn5dq932klr258nwl7e7g8rh77p97czjo.png)
and then take the natural log of both sides.
![ln(.3)=ln(e ^(-.0001t) )](https://img.qammunity.org/2019/formulas/mathematics/high-school/z8rxqfafnb4bcmkphu9h9rzcr3z5tmkm1g.png)
. But like I said, taking the natural log of the right side undoes Euler's number, so what the rule allows us to do is eliminate the ln and e:
![ln(.3)=-.0001t](https://img.qammunity.org/2019/formulas/mathematics/high-school/j4zws30obnudpj3r23f9j38zbz6ij8g70a.png)
. The natural log of a decimal will always be a negative number, -1.203972 to be exact.
![-1.203972=-.0001t](https://img.qammunity.org/2019/formulas/mathematics/high-school/4f6tzwoh0xkkjohr60f1mf0dsl33cqiy7p.png)
. We divide both sides by -.0001 to solve for t. We find that the hide, then, is 12,039 years old.