a hard math question?

Question

the demand for a certain mineral is increasing at the rate of 4.2% per year. that is dA/dt = .042A, where A = the amount used and t = the time in years from 1990
a) given that 300,000 tons were used in 1990, find the equation A(t)
b) when will the demand be double that of 1990

this may be easy for some so please be gracious in helping, i'm not good at math, never have been, thanks

Answer 1

since
dA/dt = .042A
by rearranging the terms we get
dA/A = .042dt
integrating on both sides we get
log A = .042t +k [[[since intyegral of dA/A is log A to base e]]]

first answer::
300,000 tons were used in 1990
since in 1990 the number of years ie t=0
and the amount used is A=300,000
so k=log 300,000

now the equation is
log A = .042t +log 300,000
ie log A - log 300,000 = .042t
ie log A/300,000= .042t
ie A/300,000=e^ .042t
ie A=300,000 * e^ .042t
A(t) means equation to be written with A on one side
and the other side variable as t
so A(t)=300,000 * e^ .042t

answer for b::
since amount has to be doubled A=600,000
so by substituting we get e^.042t =2
ie log 2 = .042t
ie t=(1000/42) log 2
so after t years from 1990 the demand is doubled
calculate t and then add it to 1990 to get the year when demand is doubled

Answer 2

About 17 years.
When all else fails, there is a quicky equation you should know that relates a reaction constant k to half-life: k = 0.693/t-1/2.
In this situation, you can use the same relation; since k is positive, the t you will find is the time to double. Doing this t(double) = 0.693/0.045

BTW The equation is A= Ao exp(kt), where k=0.042 and Ao=300,000. Note that
A/Ao= exp(kt) and if A/Ao=2, ln 2= 0.693=
kt(double) !!