Question #a1b58

1 Answer

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 #\lambda# is the growth factor and #\tau# is the doubling time.
We are given the doubling time (#\tau#) and the initial mass #m_0# and are asked to estimate #t# for #m(t)# 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(t)=75# kg.

#\tau=2 hrs \quad => \quad \lambda = 0.3465 hr^{-1}; \qquad t=?#
#m_0=60 pg = 6.0\times10^{-14} kg, \qquad m(t) = 75 kg#

# 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,

#\ln(ab)=\ln(a)+\ln(b)# and #\ln(x^y)=y\ln(x)#.

Use these to write #\ln(12.5\times10^{14})# as #\ln(12.5)+14\ln(10)#
#\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.