a) n(t)=n_{0}2^{(t/a)} now at time t=0, n=n_{0} but at t=19 years, n=2n_{0}

so

n(t=19 years) = n_{0}(2)^{(19/a)} which must be equal at that time to 2n_{0}

so n_{0}(2)^{(19/a)} = 2n_{0}^{ }so 2^{19/a} = 2 so 19/a = 1 so a = 19 years

and n(t) = n_{0} 2^{t/19}

b) n(t) = n_{0}e^{(rt)} to find what r, the coefficient of t in the exponent is,

n(t=19 years)= n_{0}e^{r(19 years)} =2n_{0}

so e^{r(19 years)} = 2

we take the NATURAL logarithm of both sides (that the logarithm for base e)

r(19) = ln(2)

r = ln(2)/19 = 0.0365 year^{-1}

so n(t) = n_{0}e^{(rt)} = n_{0}e^{,0365t}

c) Sorry I don't have the graph capability.

d) what time does it take to get for the population to reach 500,000?

You can use either of the two expressions; I'll use the one from a)

n(t) = n_{0} 2^{t/19}= 500000

n_{0}=^{ }120000

so (120000) 2^{t/19}= 500000

2^{t/19} = (500000/120000)=4.17

take the natural log of both sides

ln(2^{t/19}) = ln(4.17)

Note that ln(a^{b}) = b ln(a) so

(t/19)ln(2) = ln(4.17)

t= 19 ln(4.17)/ln(2) ≅39.1 years (as a check notice that since the population doubles every 19 years, it will quadruple in twice that time 38 years, at which time the population with be 480000, so it makes sense it would reach 500000 a relatively short time after that.