To find that value of logarithmic term u4= log16 8,
Through applying the change of base formula in the Graphing Calculator we get,
= 0.75
(4,0.75) is plotted on the plane and it is also point on the graph, f(x) = . Thus, in order for a logarithm to be a part of the sequence of the first row, its value along with the term number(n) (where the value is a y co-ordinate and (n) is the x-coordinate) must be a point on the graph f(x) =
Thus, the next two terms of the sequence have been verified both algebraically and graphically.
(B)
The sequence for the second row is given as:
Similarly , the sequence of the 2nd row has the expression:
log3n 81 (this is because, the bases (b) of each of the terms in the sequence are powers of 3 while A, which is 81, remains the same for each term)
= [Applying change of base formula]
= [Substituting 81 as a power of 3]
=
Now, the expression , should provide us with the values of each term in the sequence. Thus, to prove as the expression for the second row of sequences in the form :
un =
First term of the sequence =
= log3n 81
(Taking the first term as u1 and so on,)
L.H.S R.H.S
u1 = log31 81
u1 = 4 log3 81 = 4
L.H.S = R.H.S. Hence, it is verified that is the general expression for the 2nd row of the given sequence in the form .
Similarly, other terms in the sequence were also verified by using the TI-83 where:
U2: log981 (2)
U3: = log2781 (1.33)
U4: : = log8181 (1)
U5: : = log24381 (0.8)
U6: : = log72981 (0.66)
The graph above is a result of plotting f(x)= and plotting the values of each term in the sequence.
(C)
Now, the sequence of the 3rd row has the expression:
Log5n 25 (this is because, the bases (b) of each of the terms in the sequence are powers of 5 while A, which is 25, remains the same for each term)
= [Applying change of base formula]
= [Substituting 25 as a power of 5]
=
Now, the expression , should provide us with the values of each term in the sequence. Thus, to prove as the expression for the second row of sequences in the form :
un =
First term of the sequence =
= log5n 25
(Taking the first term as u1 and so on,)
L.H.S R.H.S
u1 = log51 25
u1 = 2 log525 = 2
Similarly, other terms in the sequence were also verified by using the TI-83 where:
U2: log25 25 (1)
U3: = log12525 (0.66)
U4: : = log62525 (0.5)
U5: : = log312525 (0.4)
U6: : = log1562525 (0.33)
L.H.S = R.H.S. Hence, Verified that is the general expression for the 3rd row of the given sequence in the form .
(D)
Also given, is the sequence for the mth term:
Logm mk, logm2 mk , logm3 mk, logm4mk, …
Thus, we can establish the expression of this sequence to be:
Logmn mk
=
=
Therefore, the general expression (which is in the form ) can be applied to all the given sequences. We can apply this general expression to verify the successive terms of each sequence.
Below is the general graph for the functions f(x)= , g(x) = , h(x) =
The above graph shows all three expressions from the given sequences plotted together with the values of each term in their respective sequence. Now, if we observe the graph above very carefully, we can see that the points are on a straight line, but most importantly, we can also observe that each point is equidistant from the next for each term. If this is proved to be true, one can graph even more functions to add to the given set of sequences.
To prove, the distance between points a and b, and b and c is the same for each term on the x axis.
That is, C-A= B-C
Now, by extracting the values from the same table given above earlier, we can solve the mentioned problem .
Verification : C-A = B-C
[Calculations]
Term 1: 3-2 = 4-3 = 1
Term 2: 1.5-1.0=2.0-1.5= 0.50
Term 3: 1.0-0.7 (rounded from 0.66)=1.3-1.0 = 0.30
Term4: 0.75-0.50= 1.00-0.75 = 0.25
Term 5: 0.60-0.40=0.80-0.60 = 0.20
Term 6: 0.50-0.33= 0.67-0.50 = 0.17
Hence proved, that each point is equidistant to the next for each point on the x-axis (the term of the sequence). Now that the common difference has been found, we can plot more points on the graph according to the difference in height. Below is a graph with differences by adding the differences graphically, starting from the topmost graph y = .
Full view of graph
Thus, as seen in the graphing program, a new graph has been created on the bases of the common difference found between the distances between the points of the previous graphs. The new graph has the equation of y= in the form
PART 2.
All calculations have been made by the TI-83 graphing calculator and this has been mentioned where some calculations have not been shown. The aim is to obtain two more examples fitting the pattern of the sequence given below, in addition to describing how to obtain the third answer in each row from the first two answers.
The given sequence is
Log4 64, log6 64, log32 64
Log7 49, log49 49, log343 49
Log 125, log, log 125
Log8 512, log2 512, log16 512
(A)
The terms in the sequence above are in the form of
Now, in order to obtain the third answer in each row from the first two answers, one must multiply the bases of the first two terms in the sequence to obtain the base of the third term. For example, in the first sequence, we multiply the bases 4 and 6 to obtain 32 while the logarithm of A remains constant for that sequence. Thus making the third term log32 64. This is true for all given sequences and has been verified with the use of the TI-83 calculator.
Therefore, if U1=
And U2 =
Then U3=
If = c
And, ;
In terms of c and d we can express the value of as
To prove value of =
In the second row we have, log7 49 and log49 49, where c=2 and d= 1. And a=7 and b=49
The value of the third term () is =
To verify if the value of the third term
Hence, by evaluating the third term through the graphing calculator while using the change of base formula, it is confirmed that the value of .
Similarly, all the evaluated values of last term in each sequence were the same as the values of the first two terms when they were evaluated with the formula with the help of the graphing calculator.
(B)
Let us take the pattern above while regarding base as 5 from the bases of the first column. We get:
4
7
5
8
We see that there is a difference of 3 between the 2 pairs of terms. We observe that after adding 3 to U1 in order to get U2, we subtract 2 from U2 to get U3 and so on. Therefore, the next two terms of this pattern will be 6 and 9.
Now, lets consider the pattern of the second column. This time, not the base of the logarithm, but the value of the logarithm will be taken. Thus for the second column after calculating the values of the terms, we obtain the values: 3 (log864), 1(log4949), -1(Log 125) and 9 (log2512) respectively.
Similarly, let us get the values of the logarithms in the first column. We obtain 3(log464), 2(log749),-3(Log 125) and 3 (log8512) .
Now, we have found out the all values of c and d for in the sequences given to us. The grid looks like:
Column 1(c) Column 2(d)
3 2
2 1
-3 -1
3 9
However, we can now get back to our formula we can now add values to the third column by applying this formula to each of the values in the first and second columns respectively .(with the help of the graphing calculator)
Therefore, Column 3
1.5
0.66
-.075
-4.5
Thus, by verifying with a graphing calculator, these values above are equal to the values of the logarithmic terms given in the sequences.
Therefore, following the pattern, another example which will fit the pattern above is.
Log6 36, log36 36, log 216 36
Log 729, log729 729, log 81 729
This is proved by :
Log6 36= 2
log36 36= 1
log 216 36=
according to the formula we get
Hence, since is equal to the value of the last term, the first sequence is verified.
For the second sequence,
We obtain Log 729 = -3
log729 729 = 1
log 81 729 = 1.5
according to the formula we get or 1.5
Hence, since is equal to the value of the last term, the second sequence is also verified.
The limitations of a, b and x is that we can only take positive values of a and b because logarithms of negative values don’t exist. Therfore, a, b and x can’t be negatives even though it is given that p,q, Therfore, p,q are only confined to p,q[+]