# Question a1b58

Dec 19, 2014

Since the yeast divides every two hours, its population (and hence yeast mass) doubles every two hours. Because the doubling time is a constant it is exponential growth.

The mass of yeast grows exponentially with time as :
m(t)=m_o\exp(\lambdat); \qquad => \lambda = \ln(2)/\tau,
where $\setminus \lambda$ is the growth factor and $\setminus \tau$ is the doubling time.
We are given the doubling time ($\setminus \tau$) and the initial mass ${m}_{0}$ and are asked to estimate $t$ for $m \left(t\right)$ to reach the value of the mass of a typical human. Let us assume $75$ kg as the mass of a typical human, so $m \left(t\right) = 75$ kg.

\tau=2 hrs \quad => \quad \lambda = 0.3465 hr^{-1}; \qquad t=?
${m}_{0} = 60 p g = 6.0 \setminus \times {10}^{- 14} k g , \setminus q \quad m \left(t\right) = 75 k g$

 m(t) = m_o\exp(\lambda t); \quad => \quad t=\ln((m(t))/m_o)/\lambda = (\ln((m(t))/m_o))/(\ln(2))\tau

If your calculator struggles to evaluate the logarithm of such a large number you may have to assist it. Remember the following results,

$\setminus \ln \left(a b\right) = \setminus \ln \left(a\right) + \setminus \ln \left(b\right)$ and $\setminus \ln \left({x}^{y}\right) = y \setminus \ln \left(x\right)$.

Use these to write $\setminus \ln \left(12.5 \setminus \times {10}^{14}\right)$ as $\setminus \ln \left(12.5\right) + 14 \setminus \ln \left(10\right)$
\ln((m(t))/m_o)=34.7619; \qquad t=(34.7619)/(\ln(2)) \times 2 hrs=100.3# hrs.

So it takes $100.3$ hrs, which is approximately, $4$ days and $4$ hours for the yeast to grow to the mass of a typical human.