- Level: International Baccalaureate
- Subject: Maths
- Document length: 1188 words
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. ...read more.
Middle
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 noticed a pattern : the exponent of the base of the answer in the third column is simply the sum of the exponents of the bases in the first two columns when each base is in its lowest prime form. ...read more.
Conclusion
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=c would 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. ?? ?? ?? ?? ...read more.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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