IB MATH TYPE I Portfolio - LOGARITHM BASES
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Introduction
LOGARITHM BASES
Purpose: My purpose for this investigation was to find the general statement that expresses logabx, in the terms of c and d. I was able to achieve this goal through the process of finding the expression for the nth term in various different sequences.
In the beginning stages of my investigation I came across the sequence of
Log28, Log48, Log88, Log168, Log328.......
While I was looking at this sequence I came to the realization that the base of the log form was increasing at a constant pattern. I realized that the base of 2 was being raised to the power of the term number. Therefore the second term has a base of 4 due to the fact that 2 raised to the power of 2(term number) equals 4. Even though the base changes the 8 stays constant and this also means that I would be able to continue the pattern.
6th term = log(2˄6)8 7th term = log(2˄7)8
= log648 = log1288
After continuing the pattern I realized the next step would be to solve this problem and discover what exponent we need to raise to base to, in order to achieve the answer of 8. So I decided to change my pattern from log form into exponential form.
2x=8, 4x=8, 8x=8, 16x=8, 32x=8, 64x=8, 128x=8
After successfully converting into exponential form I decided to take this a step further and make all the bases the same throughout the sequence.
2x=23,22x=23, 23x=23, 24x=23, 2
Middle
=4 ∴ =1 ∴
34=81 811=81
81=81 81=81
To further investigate expressions for different sequences will considered also consider these last sequences..
Ex. log525, log2525, log12525, log62525, log312525, log1562525
Ex. logmmk, log(m^2)mk, log(m˄3)mk, log(m˄4)mk, log(m˄5)mk, log(m˄6)mk
Observing the previous sequences I will follow the same pattern and try to come up with and expression in the same form I have used for the pervious sequences.
5x=25, 25x=25, 125x=25, 625x=25, 3125x=25, 15625x=25
5nx=52, 5nx=52, 5nx=52, 5nx=52, 5nx=52, 5nx=52
Therefore the expression is nx=2 or x=2/n
Now I will check the validity of my expression with log2525, log62525
x=2/n x=2/n
x=2/2 x=2/4
x=1 x=0.5
Now I will check my answer
Log2525 log62525
=log25 =log25
Log25 log625
=1 ∴ =0.5 ∴
251=25 6250.5=25
25=25 25=25
m1(x)=mk, m2(x)=mk, m3(x)=mk, m4(x)=mk, m5(x)=mk, m6(x)=mk
mnx=mk, mnx=mk, mnx=mk, mnx=mk, mnx=mk, mnx=mk,
Therefore the expression is nx=k or x=k/n
Continuing my investigation I decided to look at other sequences however this time instead of
finding an expression that represents the sequence I will calculate the following sequences by giving my answer in p/q form.
Log464, log864, log3264
Log4
Conclusion
ax = c, logbx = d, logabx . This general statement does not work for all sequences for example it will work for the sequence of log28, log48, log88.
...read more.
Now that I have come up with a successful general statement I would like to recall how I was initially able to come up with a general statement. When I look back at the first set of example I did with the longer sequences it helped me understand the way an expression could be formed in the format of p/q. It just gave me an overall understanding of how logs work and how you can convert logs into exponential which can be beneficial when solving for logs. Later when I moved on to my second set of example with the set of three terms in each sequences really helped me with finding the final general statement. My previous experience of forming expressions in the p/q form really helped me with these sequences. When I solved each term in all the sequences and put the answers in fraction form it became very obvious how I could use the first and second term answer to arrive at the third term answer. Seeing all the term answers side by side really helped me connect all the dots together. I was able to tell if I multiplied the two answers then divided them by the sum of the two terms I will be able to arrive at the answer.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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