# Math Portfolio 1

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Introduction

ST. ROBERT CATHOLIC HIGH SCHOOL

PORTFOLIO ASSIGNMENT 1

## OCTOBER 2009

LOGARITHM BASES SL TYPE I

This task consists of two parts. While both parts consider logarithms with different bases of the same argument, these parts are not necessarily directly related to each other.

PART 1 Exploring

Determine the numerical values of the following sequences. Explain how you got these values. Justify your answers using technology.

Use of Technology

log28 = x log48 = x log88 = x log168 = x log328 = x

2x = 8 4x = 8 8x = 8 16x = 8 32x = 8

2x = 23 22x = 23 23x = 23 24x = 23 (2)5x = 23

x = 3 2x = 3 3x = 3 4x = 3 5x = 3

log381 = x log981 = x log2781 = x log8181 = x

3x = 81 9x = 81 27x = 81 81x = 81

3x = 34 32x = 34 33x = 34 34x = 34

x = 4 x = 2

log525 = x log2525 = x log12525 = x log62525 = x

5x = 25 25x = 25 125x = 25 625x = 25

5x = 52 52x = 52 53x = 52 54x = 52

x = 2 x = 1

Write the next two terms in each of these sequences, in both logarithmic and numerical forms. Explain how you got these values.

The next two terms for the first sequence is based off this equation: where n represents the term number. In this case the next two term numbers are 6 and 7.

and

= log648 = log1288

log648 = x log1288 = x

64x = 8 128x = 8

26x = 23 2

Middle

Now consider the general sequence,

Determine the values of the first five terms. Explain how you got these values.

logmmk = x

mx = mk m2x = mk m3x = mk m4x = mk m5x = mk

x = k 2x = k 3x = k 4x = k 5x = k

Write the nth term of this sequence, in logarithmic form and in the form . Explain how you got this value.

The nth term of this sequence is:

mnx = mk

nx = k

What must be the relationship between the argument and first base if each term in the sequence is to have the form ?

The relationship is that both the sequences have the form where mq equals the base and mp equals the resultant number of the logarithm. When both the base and resultant number are turned into the same base m to the power of p and q, the answer will be in the form of .

PART 2 Exploring

Determine the numerical values of the following sequences. Explain how you got these values. Justify your answers using technology.

Use of Technology

log464 = x log864 = x log3264 = x

4x = 64 8x = 64 32x = 64

4x = 43 23x = 43 25x = 43

2

Conclusion

6x = 216 36x = 216 216x = 216

6x = 63 62x = 63 63x = 63

x = 3 2x = 3 3x = 3

x = 1

First answer: 3

Second answer:

Third answer: 1

Now consider the general case of .

Let

Determine the general equation for in terms of and .

The general equation for logab x in terms of c and d is:

Proof:

logax = c, ac = x, a =

logbx = d, bd = x, b =

logabx = y

logabx = y

(ab)y = x

Test the validity of this equation using other values of , , and .

Testing the equation:

log11 1331, log121 1331, log13311331

log111331 = x log1211331 = x log13311331 = x

11x = 1331 121x = 1331 1331x = 1331

11x = 113 112x = 113 113x = 113

x = 3 2x = 3 3x = 3

x = 1

First answer: 3

Second answer: 3/2

Third answer: 1

Discuss the scope/limitations of , , and .

Limitations:

a, b, ab:

- has to be always bigger than 0 but not equal to 1 because we cannot evaluate the logarithm of a negative base
- cannot equal to 1. Using the change of base formula, the logarithm of the bases is the denominator of the equation,. When y equals to 1, the denominator is 0 as the logarithm of 1 is 0. Any number divided by 0 is undefined.

x:

- is bigger than 0 because we cannot evaluate the logarithm of a negative number

Use the rules of logarithms to justify this equation.

Given: logax = c, logbx = d, logabx = ?

(equation 1)

(equation 2)

(from equation 1 and 2)

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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