At a certain time, bacteria in a culture has 115,200 bacteria. Four hours later, the culture has 1,440,000 bacteria. How many bacteria will there be in the culture in 15 hours?

I would call the number of bacteria #y# and write the exponential function:
#y=Ae^(kt)#
with #t# the time and #A and k# two constants.

With your numerical data we have:

At #t=t_1#

#115,200=Ae^(kt_1)#

At #t=t_2=t_1+4# (4 hours later):

#1,440,000=Ae^(k(t_1+4))#
that can be written as:
#1,440,000=Ae^(kt_1)e^(4k)#

we can substitute the first equation: #115,200=Ae^(kt_1)# into the second to get:
#1,440,000=color(red)(115,200)e^(4k)#
solve it for #k#:
#e^(4k)=(1,440,000)/(115,200)=12.5#
using natural logs on both sides we get:
#4k=ln(12.5)#
#k=1/4ln(12.5)=0.63143#

At #t=t_3=t_1+15# (15 hours later)

#y=Ae^(k(t_1+15))#
that can be written as:
#y=Ae^(kt_1)e^(15k)#
again we substitute the first equation and the value of #k# found previously:
#y=115,200*e^(15*0.63143)=1,495,770,721#

Formula for exponential growth is
#y(t) = a * e^(kt)#

Where y(t) = Number at time "t" , a = Number at start.
k = rate of growth (when >0) or decay (when <0) ,t = time

#1440000 = 115200 * e^(k*4) or e^(4k)=1440000/115200=12.5# Taking log on both sides we get #k*4= ln(12.5) or k = ln(12.5)/4= 0.6314216#

After #15# hrs bacteria will grow to # y(15) =115200 * e^(k*15) = 115200 * e^(0.6314216 *15) = 1495770721 #

# 1,495,770,721 # bacteria will be in #15# hours. [Ans]