Ib Math-Log Investigation
LOG INVESTIGATION
1. INTRODUCTION
The following assessment aims to investigate logarithms and several different expressions. The following sequences (from now on referred to as P roblem1 ) is in the form of an = logmn mk , where n represents the term number and an represents the given answer.
1. a1 = log2 8, a2 = log4 8, a3 = log8 8, a4 = log16 8, a5 = log32 8, ...
2. a1 = log3 81, a2 = log9 81, a3 = log27 81, a4 = log81 81, ...
3. a1 = log5 25, a2 = log25 25, a3 = log125 25, a4 = log625 25, ...
The next set of sequences (from now on referred to as
P roblem2 ) is written as loga x = c and logb x = d, in which x is constant, a & b are different bases, and c & d are the given outcomes.
1. log4 64, log8 64, log32 64
2. log7 49, log49 49, log343 49
1
1
1
3. log 5 125, log 125 125, log 625 125
4. log8 512, log2 512, log16 512
In this assessment, I will find the next two terms in each sequence of P roblem1 , and ultimately find an expression that follows the sequence. In P roblem2 , I will describe how I got the third answer in each sequence, and find a general term that expresses logab x, in terms of c & d.
I will further test the validity of this statement through various methods, while discussing the scopes and/or limitations of a, b, and x.
2. P ROBLEM 1
In this part of the investigation, I will find an expression for the nth term of each sequence in the form p , q where p, q ∈ Z. The sequences in P roblem1 seem to follow a similar pattern. In the first given sequence (log2n ),
3
a1 = 3, a2 = 3 , a3 = 1, a4 = 3 , and a5 = 5 (re2
4
fer to Figure 1). If we change each log into the form mk an = logmn mk , then an = mn (Figure 2). Cancel out k the m, and we have the general expression n , which is indeed in the form p . q Now, to make sure this expression worked, I tested by using it to find the following two terms. Using the calculator, I found that log64 8 = 1 . Using the expression
2
1
Fig. 1.— Using the TI Nspire
1. INTRODUCTION
The following assessment aims to investigate logarithms and several different expressions. The following sequences (from now on referred to as P roblem1 ) is in the form of an = logmn mk , where n represents the term number and an represents the given answer.
1. a1 = log2 8, a2 = log4 8, a3 = log8 8, a4 = log16 8, a5 = log32 8, ...
2. a1 = log3 81, a2 = log9 81, a3 = log27 81, a4 = log81 81, ...
3. a1 = log5 25, a2 = log25 25, a3 = log125 25, a4 = log625 25, ...
The next set of sequences (from now on referred to as
P roblem2 ) is written as loga x = c and logb x = d, in which x is constant, a & b are different bases, and c & d are the given outcomes.
1. log4 64, log8 64, log32 64
2. log7 49, log49 49, log343 49
1
1
1
3. log 5 125, log 125 125, log 625 125
4. log8 512, log2 512, log16 512
In this assessment, I will find the next two terms in each sequence of P roblem1 , and ultimately find an expression that follows the sequence. In P roblem2 , I will describe how I got the third answer in each sequence, and find a general term that expresses logab x, in terms of c & d.
I will further test the validity of this statement through various methods, while discussing the scopes and/or limitations of a, b, and x.
2. P ROBLEM 1
In this part of the investigation, I will find an expression for the nth term of each sequence in the form p , q where p, q ∈ Z. The sequences in P roblem1 seem to follow a similar pattern. In the first given sequence (log2n ),
3
a1 = 3, a2 = 3 , a3 = 1, a4 = 3 , and a5 = 5 (re2
4
fer to Figure 1). If we change each log into the form mk an = logmn mk , then an = mn (Figure 2). Cancel out k the m, and we have the general expression n , which is indeed in the form p . q Now, to make sure this expression worked, I tested by using it to find the following two terms. Using the calculator, I found that log64 8 = 1 . Using the expression
2
1
Fig. 1.— Using the TI Nspire