SL Maths Portfolio I
8th August 2008.
Logarithm Bases.
Portfolio Type I is based primarily on logarithms, sequences and series. It first presents 3 sequences of logarithms of approximately 4-5 terms. I was first asked to determine the next two terms of each sequence (namely, terms 5 and 6, or 6 and 7 in the sequence ones case). Looking at each sequence, I first noticed that the base of each logarithm was simply the first base, which I called m1, to the power of the term trying to be determined. I concluded that the next two terms of the sequences could be found through mn = m1n , where mn is the unknown base, n is the term and m1 is the first base of the sequence.
E.g. For the first sequence, to find the 6th term:
m6 = 2(6)
m6 = 64
The following table shows the 3 sequences, including the last two terms that needed to be determined.
Term
2
3
4
5
6
7
Logarithm
log28
log48
log88
log168
log328
log648
log1288
log381
log981
log2781
log8181
log24381
log72981
log525
log2525
log12525
log62525
log312525
log1562525
My next step was to find an expression of the nth term. To find an expression of the nth term, I firstly converted the logarithms into base 10 so it may be solved on my graphics display calculator - a TI-84 Plus. I then converted the answers into fractions as shown below:
Term
2
3
4
5
6
7
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
For each sequence a pattern can be seen. In the 1st sequence, the numerator is seen to be a constant integer for each term. The denominator on the other hand is seen to be the number of the term. ...
This is a preview of the whole essay
7
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
For each sequence a pattern can be seen. In the 1st sequence, the numerator is seen to be a constant integer for each term. The denominator on the other hand is seen to be the number of the term. This can be seen in the following two sequences. Using this information I was able to determine that the formula for the 1st sequence, in terms of , is y=, where p is U1 , or first number of each sequence and q is represented by n which is the term.
The formulas for the 2nd and 3rd sequences are
y= and y= respectively. To justify my claim through technology, I used a program called Autograph 3.20.
I first entered the three expressions I had found and got the above graph.
I next entered the information in the previous table for the first sequence as well as including y= in the graph. I obtained the graph seen on the left.
The dots are the plot points of the information I manually entered into the XY data set. As seen, the equation y= cuts through each plot point, which means that it is possible to also find the 6 and 7th terms of sequence 1, as well as any other term.
Following this, I was then asked to convert the following logarithm sequences into the form of.
2
3
Logarithm 1
log464
log864
log3264
Logarithm 2
log749
log4949
log34349
Logarithm 3
Logarithm 4
log8512
log2512
log16512
I was next told to describe how to obtain the third answer in each row from the first two answers and then to create two more examples that showed the pattern as well as to find the general statement that expresses logabx in terms of c and d when logax = c and logbx = d.
Firstly, obtaining the third answer, in logarithm form, can be done by multiplying the bases of the first two logarithms. For example, the bases of log464 and log864 will give the base of the third logarithm which is 32.
On the other hand, obtaining the third answer when the first two answers are in form can be better explained through the general statement that expresses logabx.
When looking at the form, I noticed a pattern could be seen - the product of the numerators of the first two logarithms is equal to the numerator of the third. As well as this, the sum of these two numerators is equivalent to the denominator of the third. Through this information I was able to come to the conclusion that the general statement is:
Next I created two more examples that fit the described pattern and also used these test the validity of my general statement.
Two Examples that fit the pattern:
Logarithm
log39
log99
log279
Logarithm
log216
log1616
log3216
The next step was to discuss the scope and/or the limitations of a, b, and x in my general statement. To do this I tested my general statement by using different sets of values for a, b, and x.
I decided that I will first start with x, first choosing negative values before moving to positive integers whilst having different values for a and b. This is shown in the following table, showing logarithm form and calculated figure on the GDC:
x-value
logax
logbx
logabx
-2
log2(-2)
log3(-2)
Log6(-2)
ERROR
ERROR
ERROR
-1
log2(-1)
log3(-1)
log6(-1)
ERROR
ERROR
ERROR
0
log20
log30
log60
ERROR
ERROR
ERROR
log21
log31
log61
0
0
0
2
log22
log32
log62
0.63
0.39
As you can see, x?cannot be equal to or less than 0. This is further reinforced by the following graph. As seen, the graph does not cut the y-axis, forming a vertical asymptote and so x is never equal to 0 or any negative integer.
a-value
logax
logbx
logabx
-2
log-22
log32
log-62
ERROR
0.63
ERROR
-1
log-12
log32
log-32
ERROR
0.63
ERROR
0
log02
log32
log02
ERROR
0.63
ERROR
log12
log32
log32
ERROR
0.63
0.63
2
log22
log32
log62
0.63
0.39
I repeated the above table for the a-values. With values of a, it can be seen that it cannot be equal to or less than 0, much like x. This is because the base of the third logarithm is the product of the first two bases and therefore if zero were the base for one of the first two logarithms, the third logarithm would have a base of zero. This also applies to negative numbers and so therefore a>0.
The findings for a values could also be applied to the b values as the base of the third logarithm would be zero or a negative number if either of the first two logarithms were to have a base of zero or negative integer. Therefore b>0.
As well as this, I also considered "what if c+d=0?", since 0 as the denominator of the formula would then make it undefined. For a denominator of 0 to be possible, it would mean that c and d would have to be opposites, for example c=8 and d=-8 hence c+d=0.
Logarithm
log39
log19
2
-2
ERROR
As you can see, an error occurred whilst trying to calculate the third logarithm. This is due to the fact that it could not divide by 0. From this it can be determined that a ? and b?.
Overall, the limitations of my general statement, , are as follows:
x>0 a>0 b>0 a ? b?.
As stated previously, my general statement is and has been proven to work under the conditions of the limitations stated before. I obtained this statement by first looking out for patterns created by sequences in both their logarithm and , testing to see if the patterns were shared with the other sequences and then creating a statement that would allow me to express it.
Maths SL
