According to one source, the amount of data passing through mobile pho

Step 1
Given: The amount of data passing through mobile phone networks doubles each year.
Step 2
a) We have given that the amount of data passing through mobile phone networks doubles each year.
That is if we assume the initial amount of data is ${\displaystyle A_0}$ and A(t) is amount of data after t years, then after one year the amount of data becomes ${\displaystyle A\left(1\right)=2A_0}$.
After 2 years the amount of data becomes ${\displaystyle A\left(2\right)=2A\left(1\right)=2\left(2A_0\right)=4A_0}$.
After 3 years the amount of data becomes ${\displaystyle A\left(3\right)=2A\left(2\right)=2\left(4A_0\right)=8A_0}$.
Continuing in this way after n years we get, ${\displaystyle A\left(n\right)=2A(n-1)}$.
Therefore, as time increases by 1 year the amount of data is obtained by multiplying the previous amount by 2.
That is the amount of data growing by constant multiples with time. Therefore, the amount of data passing through mobile phone networks is an exponential function of time.
The exponential function is of the form ${\displaystyle A\left(t\right)=A_0{e}^{kt}}$ where A(t) is amount of data after t years, ${\displaystyle A_0}$ is the initial amount of data and k is a growth factor.
Now, we have to find the value of k.
After 1 year the amount of data becomes ${\displaystyle A\left(1\right)=2A_0}$.
By using the exponential function we get,
${\displaystyle \Rightarrow 2A_0=A_0{e}^{k\left(1\right)}}$
${\displaystyle \Rightarrow 2=e^k}$
${\displaystyle \Rightarrow k=\ln \left(2\right)\approx 0.6931}$
Therefore, the amount of data grows by a factor of ${\displaystyle \ln \left(2\right)\approx 0.6931}$ each year, this is an exponential function because the amount is growing by 2(twice) multiples.
Step 3
b) To find: The exponential function using ${\displaystyle D_0}$ as the initial amount of data and D as the amount after t years.
Use ${\displaystyle D_0}$ as the initial amount of data and find a formula that gives the data D as an exponential function of the time t in years.
We get,
${\displaystyle D\left(t\right)=D_0{e}^{\ln \left(2\right)t}=D_0{\left(e^{\ln \left(2\right)}\right)}^t=D_0{\left(2\right)}^t}$
Therefore, the exponential function is ${\displaystyle D\left(t\right)=2^t{D}_0}$.
Step 4
c) To find: How long will it be before the amount of data is 100 times its initial value if the trend continues.
Use ${\displaystyle D\left(t\right)=100D_0}$ in the exponential function from part (b) we get,
${\displaystyle 100D_0=2^t{D}_0}$
Solve for t,
${\displaystyle \Rightarrow 100=2^t}$
Apply log on both sides,
${\displaystyle \Rightarrow \log \left(100\right)=\log \left(2^t\right)}$
${\displaystyle \Rightarrow 2=t\log \left(2\right)}$
${\displaystyle \Rightarrow t=\frac{2}{\log \left(2\right)}\approx 6.644}$
Therefore, it will take 6.644 years to make the amount of data 100 times its initial value.