# Thread: Expression for population in h hours

1. ## Expression for population in h hours

Hi!

Suppose that a population insects doubles every 5 hours. If you start with 6000 insects:

Find an expression for the number of insects present in $\displaystyle h$ hours.

$\displaystyle n(h) = n_0 e^\text{rh}$

$\displaystyle n(h) = 6000e^{5\text{h}}$

$\displaystyle \frac{n(h)}{6000} = e^{5\text{h}}$

$\displaystyle \ln\left(\frac{n(h)}{6000}\right) = 5\text{h}$

$\displaystyle \frac15\ln\left(\frac{n(h)}{6000}\right) = h$

2. ## Re: Expression for population in h hours

If you start with 6000, and every 5 hours it doubles, then when h = 5, n = 12 000. So

\displaystyle \displaystyle \begin{align*} 12\,000 &= 6000 \, e^{5r} \\ 2 &= e^{5r} \\ \ln{(2)} &= 5r \\ \frac{1}{5}\ln{(2)} &= r \end{align*}

So the equation which gives you the number of insects present after h hours is

\displaystyle \displaystyle \begin{align*} n(h) &= 6000 \, e^{\frac{h\ln{(2)}}{5}} \\ &= 6000 \, e^{\ln{\left( 2^{\frac{h}{5}} \right)}} \\ &= 6000 \left( 2^{\frac{h}{5}} \right) \end{align*}.