How should I start this problem?

There are many different models for handling the growth of a population. If the model dictates that the population grows by a fixed percentage of the population each year, this is known as an exponential growth pattern and has a formula of this form:

#y = A*e^(kt)#

In this formula, #A# represents the initial population (at time #t=0#). The constant #k# is a measure of growth (if positive) or decay (if negative), and the value of #k# determines the population change.

In your first case, we have 245 owls initially, and we are told that each year adds 3% of the population in growth. We've been given specifically the two constants right in the problem:

#A = 245, k = 0.03#
#:. y=245e^(0.03t)color(white)("aaaaaaaaaa")[A]#

Why does #k=0.03#? Recall that 3%, written in decimal form, is 0.03.

In the second case, we are still given the initial population (#A=63#), but this time we're not told specifically what % of the population is growth each year #t#. Instead, we're told that the population will double in #t=10# years. This means we must solve for #k# in order to have a function of #t#:

#y = A*e^(kt)#
#y = 63e^(kt)#

With the population known to be 126 at #t=10#:

#126 = 63e^(k*10)#
#2 = e^(10k)#
#ln 2 = ln (e^(10k)) #
#ln 2 = 10k#
#k = (ln 2)/(10)#
# :. y=63e^((t*ln 2)//10)color(white)("aaaaaaaa")[B]#

To find when they are at equal populations, set the equations equal to each other and solve for #t#:

#245e^(0.03t)=63e^((t*ln 2)//10)#

This is not an "easy" equation to solve for #t# by hand and it is likely you would use a Computer Algebra System or graphing calculator to solve. However, we can approximate with some decimals and #ln#s:

#245/63 = (e^((t*ln 2)//10))/(e^(0.03t))#
#3.888 = e^(0.069t-0.03t)#
#3.888 = e^(0.039t)#
#ln 3.888 = ln e^(0.039t)#
#1.358=0.039t#
#t = 34.82#

It will be about 34.82 years for equal populations.