Ib Math-Log Investigation

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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 software for Windows, I was able to confirm that the answers to the given logs were correct. k
n,

substitute 3 for k and 6 (as it is the sixth term) for
k
n, and we find that n does in fact equal 1 . Repeat the
2
process .for the next set of numbers, and we get that
log128 8 = 0.4285. Substituting 3 into k and 7 into n, we
.
get that 3 = 0.4285 as well (Figure 3). Using this ex7
pression, I tested the other sequences and confirmed that
it does still work (Figure 4).To further demonstrate that
k
3
the expression n works, I graphed k (Figure 5) which
agrees with the results in Figure 1.
k
Since the expression n is a fraction, and anything divided by zero is U N DEF IN ED, then n = 0; which makes sense since there’s no such thing as the 0th term
of a sequence. Not only that, but since log is the inverse
function of an exponential function, and the exponential
function can never be negative, m becomes restricted to
positive integers only.
k
In order to make sure that n worked for all logarithms
k
in the form an = logmn m , I decided to include several
different logarithms. The first one was log7 343, which
equals 3 when put into a calculator. I rewrote log7 343
as log71 73 , (in which 3 = k and 1 = n), used our genk
eral expression n , and found that 3 does indeed equal 3.
1
The second example I used was log144 20736, which by
checking in my calculator I know equals 2. Rewritten as
log122 124 , (k = 4 & n = 2), we find that our expression
k4
n ( 2 ) stil works. (All of this is confirmed in Figure 6). In conclusion, when solving logarithms in the form
k
an = logmn mk , an will always equal n as long as p, q
∈ Z, n = 0 and m is a positive integer.
3. P ROBLEM 2

In this section of the assessment, I analyzed the set of
sequences in P roblem2 in hopes of figuring out a recursive formula that would allow me to get the third answer

Fig. 2.— This demonstrates how an =

mk
.
mn

Fig. 6.— Calculator check justifying my answers.

Fig....
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