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....