Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
6. Natural Logarithms
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Exercise 64 Page 482

Practice makes perfect
a Notice that we are given the format of an exponential function for our model.
y=ae^(ct) In this function, t represents the amount of years after 2008 and y is the amount of internet users in billions. We can substitute the known values for the years 2008 and 2015 to find the parameters a and c. On 2008 there were 1.5 billions of internet users. We can represent this by using the values t=0 and y=1.5.
y=ae^(ct)
1.5=ae^(c( 0))
Solve for a
1.5=ae^0
1.5 = a(1)
1.5 = a
a =1.5
We have found the value for the parameter a. We can now use this together with the other known year to find the value for c. The other known year is 2015, on which there is expected to be 3.5 billion internet users. We can represent this by using the values t=2015-2008=7 and y=3.5.
y=ae^(ct)
3.5= 1.5e^(c( 7))
3.5=1.5e^(7c)
3.5/1.5=e^(7c)
7/3=e^(7c)
Recall that we can solve a natural base exponential equation by taking the natural logarithm at both sides.
7/3=e^(7c)
Solve for c & round to 3 decimals

ln(LHS)=ln(RHS)

ln 7/3 = ln (e^(7c))

ln(e^a) = a

ln 7/3 = 7c
ln 7/3/7 = c
c = ln 7/3/7
c=0.121042 ...
c ≈ 0.121
Therefore, by using the approximate value for c and the value we found for a before, we can finally write our exponential model equation. y= 1.5e^(0.121 t)
b We can use the equation we found in Part A. If we substitute the amount of internet users we are interested in, we can solve for the corresponding year. Let's give it a try.
y=1.5e^(0.121 t)
2 = 1.5e^(0.121 t)
Solve for t & round to 2 decimals
2/1.5 = e^(0.121 t)
4/3 = e^(0.121 t)

ln(LHS)=ln(RHS)

ln 4/3 = ln e^(0.121 t)

ln(e^a) = a

ln 4/3 = 0.121 t
ln 43/0.121 = t
t = ln 43/0.121
t = 2.377537 ...
t = 2.38
Recall that the years are being measured from 2008. Therefore, since 2008 + 2.38 = 2010.38, we can expect there to be 2 billion internet users some months after the beginning of 2010.
c We will proceed just at we did at Part B. This time we will use y = 5 to represent 5 billion users in our function.
y=1.5e^(0.121 t)
5 = 1.5e^(0.121 t)
Solve for t & round to 2 decimals
5/1.5 = e^(0.121 t)
10/3 = e^(0.121 t)

ln(LHS)=ln(RHS)

ln 10/3 = ln e^(0.121 t)

ln(e^a) = a

ln 10/3 = 0.121 t
ln 103/0.121 = t
t = ln 103/0.121
t = 9.950188 ...
t = 9.95
Then, since 2008+9.95 = 2017.95, we would expect to have 5 billion internet users near the end of 2017.
d Proceeding in a similar way as in Parts B and C, we can solve for t directly for the function found in Part A.
y=1.5e^(0.121 t)
Solve for t
y/1.5 = e^(0.121 t)

ln(LHS)=ln(RHS)

ln y/1.5 = ln e^(0.121 t)

ln(e^a) = a

ln y/1.5 = 0.121 t
ln y1.5/0.121 = t
t = ln y1.5/0.121
e In order to verify the answers in Part B and Part C, we can introduce the amount of users to find the corresponding year. The results found using this formula should be the same we obtained before.