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

Question

2761 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-20T19:44:36

Start with $Q(t) = Q(0)e^(lambdat)$
Use the fact that $(Q(t_"half-life"))/(Q(0)) = 1/2$ to find $lambda$

Start with $Q(t) = Q(0)e^(lambdat)" [1]"$

At $t_"half-life" = 1$

$(Q(1))/(Q(0)) = 1/2 = e^(lambda(1))$

Use the natural logarithm on both sides:

$ln(1/2) = ln(e^(lambda(1)))$

Flip the equation:

$lambda = ln(1/2)$

This is a better form:

$lambda = -ln(2)$

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

$Q(t) = 1500e^(-ln(2)t)$