Exponents and logarithms, multiple pages

Exponential Growth and Decay

The law of Uninhibited Growth can be termed as follows:

With $A_0$ the initial amount at the beginning ( time t=0) and the constant $k \neq 0$

$A(t)=A_0e^{kt}$

This function can be a growth or a decay depending on the value of $k$
If $k>0$ this is an exponential law or the Law of Uninhibited Growth.
If $k<0$ it is the Decay.

For a Growth of cells, we will have $k>0$

It is usually noted:

$N(t)=N_0e^{kt}$

Radioactive Decay:

It is the same formula but with $k<0$

$A(t)=A_0e^{kt}$

For the carbon dating, we refer to the half-life. This is about $5600\; years$ for the carbon 14.
While the carbon 12 won’t change, it helps to find the time when a given organism died, when compared to the amount of carbon 14.

Example:
The amount of carbon 14 found in a carbon 14 dating process is 2.15% of the initial amount.
Using the half-life of carbon 14 as 5600 years, when did the organism die?

Solution
$A(t)=A_0e^{kt}$

For $t=5600$ the value is half of the original:

$\frac{1}{2}A_0=A_0e^{5600k}$

$\frac{1}{2}=e^{5600k}$

$-\ln2=5600k$

$k=-\frac{\ln2}{5600}$
$k=-1.23776628\times 10^{-4}$
$k=-0.000123776628$

The equation is now:

$A(t)=A_0e^{-0.000123776628t}$

In this situation, $A(t)=0.0215A_0$

$0.0215A_0=A_0e^{-0.000123776628t}$

$0.0215=e^{-0.000123776628t}$
$\ln(0.0215)=-0.000123776628t$
$t=\frac{\ln(0.0215)}{-0.000123776628}$
$t \approx 31,021$ years ago.

Newton’s law of cooling:

The $u$ temperature of a heated object at a given time $t$ , with $k<0$ , $u_0$ the original temperature and $T$ the temperature of the surrounding medium.