# Logarithm Bases

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Introduction

## Logarithm Bases

In the beginning of this problem, we are given multiple sequences of logarithms, and are told to write down the next two terms of each sequence. Here is a table of the first sequence, including the next two terms and the numerical equivalence of each term:

# of Term | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Term | |||||||

Numerical Equivalence |

I created the last two terms in this sequence, terms 6 and 7, simply by doubling the base of the logarithm for each term. In more proper mathematical terms, I used the formula , where n is the number of the term. I noticed that each numerical equivalence seems to have 3 as the numerator, and the number of the term as the denominator. In this manner, I created a formula to find the numerical equivalence for the nth term of the sequence in the form

Middle

Numerical Equivalence

I noticed that in both of these sequences as well, the numerical equivalences seem to have a constant integer as the numerator, and the number of the term as the denominator. Therefore, the formula for the first sequence is , and the formula for the second sequence is .

Now, we are asked to calculate a set of many logarithms, in the form , where p and q are both integers. The set is as follows:

Logarithm | |||

Numerical Equivalence | |||

Logarithm | |||

Numerical Equivalence | |||

Logarithm | |||

Numerical Equivalence | |||

Logarithm | |||

Numerical Equivalence |

Finding the numerical equivalences of the logarithms was very easy; I simply calculated them using a graphing calculator. The challenging part arose when I was asked to find a way to obtain the third answer in each row from the first two answers in each row. I was also told to let and , and to find a general statement that expresses , in terms of c and d.

Conclusion

Logarithm | |||

Numerical Equivalence | 2 | -2 | ERROR |

This didn’t work, because c+d=0, and the general formula ended up with a 0 on the bottom. When I tried to solve the third logarithm using my calculator, it gave me an error message saying “Divide by 0”. Therefore, and .

One more thing that I thought of was that x must be greater than 0, because of the definition of a logarithm. There is no possible way to get any number to equal 0 or any negative number using only exponents. Therefore, .

I couldn’t think of any more limitations of the general statement. Here are the limitations that I came up with:

I will now repeat my general statement. If we let and , then

.

This statement only works with the limitations stated previously. It took me a lot of staring at the numbers and trying different values for a, b, and x, but eventually I discerned a pattern and arrived at my general statement.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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