• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7

# Math Portfolio 1

Extracts from this document...

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

36216 = x                log216216 = x

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

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

x = 3                          2x = 3                            3x = 3

x = 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

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.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Maths essays

1. ## Extended Essay- Math

erï¿½'' ï¿½f\ï¿½ï¿½`ï¿½ï¿½ï¿½ ï¿½xï¿½ï¿½F|"ï¿½ï¿½~ï¿½ï¿½ï¿½...-"Pï¿½ï¿½ï¿½ï¿½Jï¿½ï¿½R2ï¿½4V9+ï¿½ï¿½(c)ï¿½}ï¿½c,`ï¿½*[email protected]ï¿½yï¿½ï¿½Å¨ï¿½(tm)#ï¿½@=ï¿½Kï¿½ï¿½_}ï¿½ï¿½ï¿½#FÈ¨ï¿½nWï¿½K-.š=ï¿½LlXc="#oï¿½ '_yD y2jï¿½3*(r)Æï¿½OOn&ï¿½Utï¿½e-ï¿½Y0ï¿½(r)...(KdÔ²ï¿½ï¿½mï¿½Í³ï¿½1/2Lï¿½ï¿½hï¿½ï¿½'ï¿½\+ATï¿½& È¨(tm)ï¿½oCï¿½8~ï¿½-ï¿½ï¿½#ï¿½Tlï¿½`o !ï¿½ï¿½dV-H'ï¿½ï¿½Z'ï¿½mï¿½ï¿½Q\$kï¿½!\$"ï¿½ï¿½{q}ï¿½ 1/2ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½Z|lï¿½(ï¿½oï¿½[ï¿½ï¿½O_{ï¿½ï¿½l0jï¿½(ï¿½ï¿½=Ò­\$ ï¿½f3/4Usï¿½}ï¿½(r)]ÉB<ï¿½ï¿½={r0kï¿½ï¿½-ï¿½[email protected] ï¿½.ï¿½Iï¿½ï¿½pï¿½Oï¿½'ï¿½ï¿½~ï¿½ï¿½wï¿½IwÏ¥6Kï¿½TPï¿½T"[email protected]ï¿½ï¿½ï¿½ï¿½8+ï¿½Yï¿½Ð§ï¿½<ï¿½ï¿½ï¿½wï¿½ï¿½ï¿½ï¿½9ï¿½ @...ï¿½"ï¿½PO-^3/4--Nï¿½(c)ï¿½}Ô¿ï¿½ï¿½ï¿½ï¿½yï¿½H"-7Vï¿½e"ï¿½ï¿½ZF]1/4(tm)ï¿½]ï¿½,ï¿½r^*ï¿½PbLï¿½ï¿½-ï¿½[email protected]ï¿½ wKjï¿½((c)Z&ï¿½rï¿½ ï¿½ï¿½ï¿½ï¿½m- D È¨"o3/4jvï¿½vï¿½ï¿½)9nï¿½ï¿½-ï¿½\3/4S'ï¿½T!ï¿½ï¿½Q"...ï¿½ï¿½ï¿½ï¿½I'+'f01wï¿½ï¿½^}ï¿½ï¿½'"ï¿½BD 4ï¿½~ï¿½Fï¿½xnï¿½ï¿½}ï¿½v/ï¿½cï¿½ï¿½ï¿½vï¿½mï¿½-ï¿½ï¿½Û·ï¿½%ï¿½B& ï¿½Zï¿½-ï¿½ï¿½ï¿½'jï¿½ï¿½%0ï¿½ï¿½ï¿½ï¿½f'ï¿½3)ï¿½ï¿½Z&ï¿½~ï¿½C{}-Aï¿½ lï¿½Bï¿½ Ò£ï¿½ È¨ï¿½Mï¿½ï¿½ï¿½(tm)Pï¿½Ü¹sï¿½ï¿½ï¿½ï¿½nSg-Zï¿½*ï¿½*2jUï¿½JUfï¿½spï¿½]ï¿½M:ï¿½ï¿½ï¿½nï¿½ï¿½Jhï¿½%"@[email protected]ï¿½tï¿½ ï¿½}BHï¿½ï¿½Njjï¿½ï¿½dÔ¬WQ<r -(r)ï¿½ï¿½ï¿½ï¿½ï¿½/ï¿½ï¿½tï¿½ï¿½8 [ï¿½Y'ï¿½ï¿½ZEDï¿½ï¿½Pmï¿½(0ï¿½Fgï¿½G?ï¿½QUo)ï¿½[email protected]ï¿½ï¿½ï¿½ï¿½g(c)-ï¿½Yï¿½n.Zï¿½ï¿½Tï¿½ï¿½@Ldï¿½bk{ï¿½ï¿½ï¿½iï¿½&>)é¬7nÚ´iï¿½Uï¿½'E (ï¿½ï¿½XÒ:ï¿½wp4ï¿½Bï¿½ï¿½Zï¿½W"Sj' ï¿½V;C'K`ï¿½lT^ï¿½ï¿½ï¿½ï¿½ Uï¿½ï¿½ï¿½ï¿½Z)>Jï¿½ï¿½ï¿½|?ckuï¿½8iï¿½ï¿½ï¿½ï¿½cSNï¿½ï¿½}?sEï¿½.Xï¿½Ò¥ˆ ï¿½=ï¿½/Wï¿½.D Zï¿½(c)UK, ï¿½{ï¿½(tm)oï¿½bj'ï¿½ï¿½3ï¿½;ï¿½(c)|+...dï¿½ (tm)ï¿½?...y(r)HbC!5qï¿½DN0PPï¿½ ï¿½"'2j'`tï¿½ï¿½ï¿½"Wï¿½=ï¿½-#ï¿½ï¿½Å±Ç¹]ï¿½@ï¿½dï¿½bï¿½kiï¿½ï¿½ï¿½}ï¿½gï¿½&Mï¿½0ï¿½ï¿½?ï¿½yï¿½ï¿½ï¿½" ï¿½ï¿½ï¿½ï¿½Ù·oï¿½Äá¹¸k#nbï¿½3^'ï¿½Zï¿½>->Y"d ?ï¿½-_yï¿½Ì¬%Yï¿½ï¿½ï¿½,ï¿½ï¿½(c)ï¿½ï¿½ï¿½KØ¢ï¿½ï¿½(c)yï¿½ï¿½aL3ï¿½[ Nï¿½ï¿½ï¿½Zï¿½ï¿½MWHlï¿½Hï¿½ï¿½ï¿½ï¿½Sc:cÆï¿½ï¿½[email protected]ï¿½_ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½ï¿½kï¿½ ï¿½[email protected]ï¿½nGï¿½ -giï¿½ï¿½ï¿½ï¿½/ï¿½Jyj! ï¿½V ='ï¿½e?GÇï¿½ï¿½Ngï¿½ï¿½"ï¿½6Rï¿½*ï¿½ ï¿½~fGï¿½ï¿½"ï¿½ï¿½ï¿½Zï¿½ï¿½y}zï¿½ï¿½ï¿½v ï¿½+ï¿½ï¿½nï¿½% ï¿½Zï¿½<m/Í8"ï¿½ï¿½gQï¿½1:xï¿½`-...ï¿½Ä¨È¨EEÒrbï¿½ï¿½ï¿½-3/4ï¿½%6ï¿½q8|ï¿½ï¿½ï¿½#!È¨...ï¿½ï¿½ï¿½+x}vï¿½ï¿½ï¿½`ï¿½(c)ï¿½O\ï¿½dï¿½ATuï¿½ È¨ï¿½[ï¿½ï¿½ï¿½ï¿½}4ï¿½ï¿½3/4bï¿½ï¿½Gï¿½ï¿½ï¿½ï¿½~Vï¿½ï¿½D Z(ï¿½ï¿½CEï¿½ï¿½ï¿½mÚ´(c)Eï¿½...2'ï¿½Pï¿½Uï¿½ï¿½|^ï¿½Hï¿½ï¿½ ï¿½Pï¿½ï¿½[email protected]=ï¿½ï¿½ï¿½:jï¿½+jï¿½'ltï¿½ï¿½Yï¿½ï¿½&ï¿½<SuÉ¨eHÝ^ï¿½Xlï¿½ï¿½X/1/4ï¿½Bï¿½@ï¿½#ï¿½ï¿½ï¿½dï¿½Bï¿½sï¿½E\ï¿½ï¿½ Yï¿½Î#X+ ï¿½X>ï¿½ï¿½ï¿½@2j!

2. ## Math portfolio: Modeling a functional building The task is to design a roof ...

Therefore width of the cuboid = 2v = 41.56 meters The value is same as in the case of 36 meters height structure hence I can say that the width of the largest cuboid does not change in the structure's height.

1. ## Math Portfolio: trigonometry investigation (circle trig)

First, in first quadrant where (00 to 900, as well as -2700 to -3600) all values of y are positive, radius which is always positive or in other words real number, dividing the y value or opposite of that will always give positive quotient.

2. ## Logrithum bases

Example 1: Log 1/9 729, Log 1/729 729, Log 1/81 729 Logarithm 1st 2nd 3rd p/q -3 -1 -3/2 Example 2: Log 6 216 Log 216 216, Log 36 216 Logarithm 1st 2nd 3rd p/q 3 1 1.5 It seems that these two examples fit on all two patterns.

1. ## Logarithm Bases - 3 sequences and their expression in the mth term has been ...

U3: = log2781 (1.33) U4: : = log8181 (1) U5: : = log24381 (0.8) U6: : = log72981 (0.66) The graph above is a result of plotting f(x)= and plotting the values of each term in the sequence. (C) Now, the sequence of the 3rd row has the expression: Log5n 25 (this is because, the bases (b)

2. ## Math Portfolio Type II

0 10000 10 59952 1 23000 11 60026 2 45126 12 59985 3 62577 13 60008 4 58384 14 59995 5 60837 15 60002 6 59513 16 59998 7 60266 17 60000 8 59848 18 59999 9 60084 19 60000 20 59999 The above tabulated data can be represented by

1. ## IB SL Math Portfolio- Logarithm Bases

One could also use the "solver" function on a GDC calculator to check their math by simply pressing the "math" key, then pressing "solve". The original statement can be written 2x=8, but in order to input this into the solver function, it needs to be set to zero.

2. ## IB MATH TYPE I Portfolio - LOGARITHM BASES

=4 ï =1 ï 34=81 811=81 81=81 81=81 To further investigate expressions for different sequences will considered also consider these last sequences.. Ex.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to
improve your own work