Question 767990
The E notation appears to be similar to scientific notation, the E disignating an exponent on 10.  
1.38E8 seems to be a way to show 1.38x10^8.


{{{N=Ie^(kx)}}}
N=count of car owners after time x
I=initial count of car owners when x is taken at 0
k=a constant
x=time passage in years since 2010



We find k first, using the doubling time.
Let N=2I and x=6.3.
{{{2I=Ie^(k*6.3)}}}
{{{2=e^(k*6.3)}}}
{{{ln(2)=6.3*k*ln(e)=6.3*k*1=6.3*k}}}
{{{k=(1/6.3)ln(2)}}}
As a value we can just calculate {{{k=0.110}}}


EQUATION TO APPLY IS {{{highlight(N=1.38*10^8*e^(0.110*x))}}}


At x=0, year is 2010.  I=1.38*10^8
Just be aware this question has an accuracy problem, because the doubling information is based on nearest tenth of a year but year point values are only to the nearest year.  Answers can be at best, only estimations.


So with this question, in year 2020, {{{x=2020-2010=10}}} years.
{{{N=1.38*10^8*e^(0.110*10)}}}
{{{N=1.38*10^8*3.00487}}}
{{{N=4.15*10^8}}}, might be better to say {{{highlight(4.1*10^8)}}}