- Level: International Baccalaureate
- Subject: Maths
- Word count: 1188
IB SL Math Portfolio- Logarithm Bases
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Introduction
IB Math SL Portfolio Type 1
General Introduction:
In its simplest terms, a logarithm is an exponent. It is an exponent needed to produce a given number from a specific base. It is written in the form loggh=j, which denotes gh=j (g to the power of h equals j).
LOGARITHM BASES
Consider the following sequences. Write the next two terms of each sequence.
log28, log48, log88, log168, log328, ... log648, log1288, ...
log381, log981, log2781, log8181, … log24381, log72981, ...
log525, log2525, log12525, log62525, … log3,12525, log15,62525, ...
:
:
:
logmmk, logm^2mk, logm^3mk, logm^4mk, … logm^5mk, logm^6mk, …
Find an expression for the nth term of each sequence. Write your expressions in the form p/q, where p and q are integers. Justify your answers using technology.
The first sequence can be otherwise written:
log2^123, log2^223 , log2^323, log2^423, log2^523, ...
Here, I noticed a pattern: the base of each logarithm is 2n. Using this knowledge and the concepts of the change of base rule and verifying my theories with a GDC calculator, I developed the expression shown here by evaluating each term.
(Each term to be evaluated can be checked by using a GDC calculator. For example, log28
Middle
log5^152= 2/1
log5^252= 2/2
log5^352= 2/3
log5^452 = 2/4
:
:
log5^n52 = 2/n
The nth term of the third sequence can expressed as 2/n.
The same rules apply for the fourth and final sequence.
logm^1mk, logm^2mk, logm^3mk, logm^4mk, …
logm^1mk= k/1
logm^2 mk= k/2
logm^3 mk= k/3
logm^4 mk= k/4
:
:
logm^n mk= k/n
The nth term of the fourth sequence can expressed as k/n.
Now, calculate the following, giving your answers in the form p/q, where p and q are both integers.
log464, log864, log3264 3/1, 2/1, 6/5
log749, log4949, log34349 2/1, 2/2, 2/3
log1/5125, log1/125125, log1/625125 3/-1, 3/-3, 3/-4
log8512, log2512, log16512 3/1, 9/1, 9/4
Describe how to obtain the third answer in each row from the first two answers.
First, I took the lowest prime base possible of each.
log2^264, log2^364, log2^564
log7^149, log7^249, log7^349
log(1/5)^1125, log(1/5)^3125, log(1/5)^4125
log2^3512, log2^1512, log2^4512
In doing this, I
Conclusion
logax=clogbx=d
log864=4 log464=2
c=4 d=3
logabx= cd/(c+d)
log3264= 6/5
or
326/5=64
Discuss the scope and/or limitations of a, b and x.
The variables a and b must be integers greater than zero and not equal to one. If a or b were equal to one, then c and d would be rendered undefined.
(For example, if a=1 and x=3, the given statement logax=cwould be undefined for c
because there is no exponent that can be raised to make one equal three. One to any power always equals one.)
This is the same for b. If it were equal to one, d would be undefined.
They cannot be negative, as this does not work with the equations.
The variable c cannot equal negative d, as this would yield a zero in the denominator of the equation, which would make it undefined.
Explain how you arrived at your general statement.
I explained it step by step above, but generally I just used my knowledge of logarithm and exponent rules to mold and simplify the information I was given. For example, I knew that logax=c was the same as writing x= ac, a point I explained in the brief introduction,which helped me not only with this portion but the portfolio in its entirety.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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