# I am going to go through some logarithm bases, by continuing some sequences and finding an equation to find the nth terms in the sequences

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Introduction

Rachel Brustad

Logarithm Bases

In the next few examples I am going to go through some logarithm bases, by continuing some sequences and finding an equation to find the nth terms in the sequences in terms offor the first part. Then for the second part, find an equation to figure out the third term in terms of from the given logarithms, after I will check my equation with several examples created by myself and check the validity of the equation, and the scope and limitations of a, b, and x.

PART 1:

With the four sequences given I will find the next two terms in each sequence then change the first three sequences into the form of the forth sequence which is necessary to find the equation to find the nth term. Finally for all four sequences I will check the validity by using my calculator and finding the 10th term and proving that my equation works.

- Log28, log48, log88, log168, log328, log648, log1288
- Log381, log981, log2781, lg8181, log24381, log72981
- Log525, log2525, log12525, log52525, log312525, log1562525

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By changing the first three sequences to the form of the forth sequence it changes the subscripts.

Middle

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= y (3n)y= 34 The 3’s will cancel out and give you ny=4 y=

Next to prove the equation I put 10 in for the nth power and put

in to my calculator which gives you .4 next I put in my calculator and also got .4 proving the equation works.

- Log525, log2525, log12525, log52525, log312525, log1562525

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= y (5n)y= 52 The 5’s will cancel out and give you ny=2 y=

Next to prove the equation using my calculator I put 10 in for n again.

Which then gives you .2 then I put in the calculator which also gives you .2 proving the equation.

- This last equation is already in the right form so all we have to do is change it to exponential form, find the equation and prove it works.

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= y (mn)y=mk The m’s will cancel out giving you ny=k y=

To prove this I again put 10 in for n.

, but then you have to divide to get the final answer the logs will cancel leaving you with. Then if you plug 10 into the equation you will also get proving that the equation works.

PART 2:

In part two I am going to calculate the answer of several logarithms in the form of

Conclusion

log4 .64= -.32= - log8 .64= -.22= - log32 .64= -.13= -

This shows that having x in decimal form will not work to fit the equation.

Log.464= -4.54= - log.864= -18.6= - log.3264= -3.65= -

This also shows that having a as a decimal will not work in the equation.

We can also put zero in as the a or x to see if that works in the equation.

log40=error log80=error log320=error

This shows that having a zero in for the x value will not work and cause an error.

Log064=error log064=error log064=error

This also shows that have zero in the for the a value will not work as well and also causes an error in the calculator.

From these examples we have proven that positive and fractions work to fit the equation but negative, decimals and zeros don’t work to fit the equation.

From all of these examples in part 1 and part 2 I have found equations to find the nth terms in the form of. Also I found an equation to find out a way to calculate the third term in the set of logarithms. I have given examples that I have made up myself and proved the validity of the problems. Finally I found the scope and limitations of the logarithms in part 2.

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

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