# How do you find the associated exponential decay or growth model: Q = 1,500 when t = 0; half-life = 1?

May 20, 2017

Start with $Q \left(t\right) = Q \left(0\right) {e}^{\lambda t}$
Use the fact that $\frac{Q \left({t}_{\text{half-life}}\right)}{Q \left(0\right)} = \frac{1}{2}$ to find $\lambda$

#### Explanation:

Start with $Q \left(t\right) = Q \left(0\right) {e}^{\lambda t} \text{ [1]}$

At ${t}_{\text{half-life}} = 1$

$\frac{Q \left(1\right)}{Q \left(0\right)} = \frac{1}{2} = {e}^{\lambda \left(1\right)}$

Use the natural logarithm on both sides:

$\ln \left(\frac{1}{2}\right) = \ln \left({e}^{\lambda \left(1\right)}\right)$

Flip the equation:

$\lambda = \ln \left(\frac{1}{2}\right)$

This is a better form:

$\lambda = - \ln \left(2\right)$

Substitute $1500$ for $Q \left(0\right)$ and $- \ln \left(2\right)$ for $\lambda$ into equation [1]:

$Q \left(t\right) = 1500 {e}^{- \ln \left(2\right) t}$