IB Mathematics SL Portfolio: Logarithmic Bases

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  • Topic: Logarithm, Negative and non-negative numbers, Integer
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  • Published : November 13, 2010
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IB Mathemetics SL Portfolio: Logarithmic Bases

In this portfolio task, I will investigate the rules of logarithms by identifying the logarithmic sequences. After identifying the pattern, I will produce a general statement which defines the sequence. I will then test the validity of my general statement by using other values. I will finally conclude the portfolio task by explaining how I arrived to my general statement and its limitations.

Consider the following sequences. Write down the next two terms of each sequence.

For the first row, the next two terms of the sequence would be log64 8 and log128 8. For the second row, the next two terms of the sequence would be log243 81 and log729 81. For the third row, the next two terms of the sequence would be log3125 25 and log15625 25.For the fourth row, the next two terms of the sequence would be logm5 mk and logm6 mk.

Find an expression for the nth term of each sequence. Write your expressions in the form p/q where p, q Z. Justify your answer using technology.

To find the nth term, I first solved the sequence on the first row. This is shown below.

Log2 8 = x Log4 8 = x Log8 8 = x Log16 8 = x Log32 8 =x ⇒2x = 8 ⇒4x = 8 ⇒8x = 8 ⇒16x = 8 ⇒32x = 8 ⇒2x = 23 ⇒(22)x = 23 ⇒x = 1 ⇒(24)x = 23 ⇒(25)x = 23 ⇒x = 3 ⇒2x = 3 ⇒4x = 3 ⇒5x = 3 ⇒x = 3/2 ⇒x = 3/4 ⇒ x= 3/5

Log64 8 = x Log128 8 = x
⇒64x = 8 ⇒128x = 8
⇒(26)x = 23 ⇒(27)x = 23
⇒6x = 3 ⇒7x = 3
⇒x = 3/6 ⇒ x = 3/7

By looking at these answers for the first sequence, I noticed the fact that the denominator of each answer shows the nth term. For example, my answer to Log2 8 (which is the first term) was 3 (or 3/1) and my answer to Log4 8 (which is the second term of the sequence) was 3/2. By looking at my answers to the first sequence, I found out that the nth term could be found by the expression 3/n.

For the second row, I solved the answers to the sequence, which is shown below.

Log3 81 = x Log9 81 = x Log27 81 = x Log81 81 = x Log243 81 = x ⇒3x = 81 ⇒9x = 81 ⇒27x = 81 ⇒81x = 81 ⇒243x = 81 ⇒3x = 34 ⇒(32)x = 34 ⇒(33)x = 34 ⇒x = 1 ⇒(35)x = 34 ⇒x = 4 ⇒2x = 4 ⇒3x = 4 ⇒5x = 4 ⇒x = 4/2 ⇒x = 4/3 ⇒x = 4/5

Log729 81 = x
⇒729x = 81
⇒(36)x = 34
⇒6x = 4
⇒x = 4/6

By looking at these answers for the second sequence, I noticed that it is similar to the first sequence. Again, the denominator of each answer shows the nth term. My answer for Log3 81 (the first term) was 4 (or 4/1) and Log9 81(the second term) was 4/2. This shows that the expression for the nth term could be found by 4/n

For the third row, I solved the answers to the sequence, which is shown below.

Log5 25 = x Log25 25 = x Log125 25 = x Log625 25 = x Log3125 25 = x ⇒5x = 25 ⇒25x = 25 ⇒125x = 25 ⇒625x = 25 ⇒3125x = 25 ⇒5x = 52 ⇒(52)x = 52 ⇒(53)x = 52 ⇒(54)x = 52 ⇒(55)x = 52 ⇒ x = 2 ⇒2x = 2 ⇒3x = 2 ⇒4x = 2 ⇒5x = 2 ⇒x = 2/2 ⇒x = 2/3 ⇒x = 2/4 ⇒x = 2/5

Log15625 25 = X
⇒15625x = 25
⇒(56)x = 52
⇒6x = 2
⇒x = 2/6

By looking at the answers for the third sequence, I found out that the denominator again shows the nth term. The expression for the nth term would be found by 2/n

For the fourth row, I solved the answers to the sequence, which is shown below

Logm mk = x Logm2 mk = x Logm3 mk = x Logm4 mk = x Logm5 mk = x ⇒mnx = mk ⇒m2x = mk ⇒m3x = mk ⇒m4x = mk ⇒m5x = mk ⇒nx = k ⇒2x = k ⇒3x = k ⇒4x = k ⇒5x = k ⇒x = k/n ⇒x = k/2 ⇒x = k/3 ⇒x = k/4 ⇒x = k/5

Logm6 mk = x
⇒m6x = mk
⇒6x = k
⇒x = k/6

By looking at the answers for the fourth sequence, the pattern...
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