Use the equation #n(t) = n_0e^(kt)# and the points #n(1) = 10# and #n(2) = 25# to write two equations:
#10 = n_0e^(k(1))" [1]"#
#25 = n_0e^(k(2))" [2]"#
We want the value of #n_0# but, unfortunately, we must solve for the value of #k#, before we can solve for the value of #n_0#.
Divide equation [2] by equation [1]:
#25/10 = n_0/n_0e^(k(2))/e^(k(1))#
The left side becomes 2.5, #n_0/n_0# becomes 1, and division of exponential functions is the same as the subtraction in the exponents:
#2.5 = e^(k(2-1)) = e^k#
Use the natural logarithm on both sides:
#ln(2.5) = ln(e^k)##
#k = ln(2.5)#
Now that we have the value of k, we can use either equation [1] or equation [2] to solve for #n_0#
#10 = n_0e^(ln(2.5)(1))#
#10 = n_0(2.5)#
#n_0 = 10/2.5#
#n_0 = 4#
