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Math Portfolio 1

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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 image00.png

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 = ximage70.pngimage79.png

     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 = 3image26.pngimage107.pngimage96.png


log381 = x        log981 = x        log2781 = x        log8181 = ximage12.pngimage21.png

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

       3x = 34                32x = 34               33x = 34                34x = 34image02.pngimage35.png

        x = 4                x = 2                                  

log525 = x        log2525 = x        log12525 = x        log62525 = ximage45.pngimage46.png

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

       5x = 52                   52x = 52                53x = 52                 54x = 52 image48.pngimage47.png

        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: image51.png where n represents the term number. In this case the next two term numbers are 6 and 7.

image52.png          and         image53.png

= log648                       = log1288

log648 = x                log1288 = x

     64x = 8                           128x = 8

     26x = 23                                  2

...read more.



Now consider the general sequence, image75.png

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 = kimage83.pngimage82.pngimage81.pngimage84.png

Write the nth term of this sequence, in logarithmic form and in the form image67.png.  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 image67.png?

The relationship is that both the sequences have the form image87.png 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  image88.pngimage88.png.

PART 2 Exploring image07.png

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 image93.png

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

        4x = 43                         23x = 43                           25x = 43image94.png


...read more.


36216 = x                log216216 = x

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

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

          x = 3                          2x = 3                            3x = 3image04.png

                      x = 1

First answer: 3

Second answer: image05.pngimage05.png

Third answer: 1


Now consider the general case of image07.png.

Let  image08.png

Determine the general equation for image09.png in terms of image10.png and image11.png.

The general equation for logab x in terms of c and d is: image13.png


logax = c, ac = x, a = image14.png

logbx = d, bd = x, b = image15.png

logabx = y

logabx = y

(ab)y = x







Test the validity of this equation using other values of image23.png, image24.png, and image25.png.

Testing the equation:image22.pngimage22.png

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

image26.pngimage26.png                            x = 1

First answer: 3

Second answer: 3/2

Third answer: 1


Discuss the scope/limitations of image23.png, image24.png, and image25.png.



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,image29.png. When y equals to 1, the denominator is 0 as the logarithm of 1 is 0. Any number divided by 0 is undefined.



  • 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 = ?

image33.png    (equation 1)image31.png


image36.png      (equation 2)image31.png



image39.png        (from equation 1 and 2)






...read more.

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