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

1. Log28, log48, log88, log168, log328, log648, log1288

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= y                (2n)y= 23    the 2’s will cancel out and give you  ny= 3               y=

Next to prove this I put

into my calculator. 10 as the nth term and that equals .3 I also put in the calculator to get .3 they equal the same which means the equation works.

2. Log381, log981, log2781, lg8181, log24381, log72981

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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.

3. 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.

4. 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

Then find an equation to find the 3rd logarithm based off of the first two in terms of c and d.  I will show that it works based off two equations that I have created myself.  Also I will see if there is any scope or limitations that won’t work in the logarithms.

We will let the first problem in each set of logarithms be logax=c, the second problem be logbx=d, and the third problem will be logabx=? because the subscript in the first two problems multiply to give you the subscript in the third problem and we are trying to figure out an equation to find what the third problem equals based of c and d in the first two problems.

1. logax=c  logbx=d  logabx=?

log464=c  log864=d  log3264=?

=3=    =2=   =1.2=

2.  logax=c  logbx=d  logabx=?

log749=c  log4949=d  log34349=?

=2=   =1=  =.66=

3. logax=c  logbx=d  logabx=?

log1/5125=c  log1/25125=d  log1/625125=?

= -3= -  = -1= -  = -.75= -

4. logax=c  logbx=d  logabx=?

log8512=c  log2512=d  log16512=?

=3=  =9=  =2.25=

Now to find an equation on how to find the third answer without having to actually calculate it in a calculator, If you look at the first two answers in forms of you can notice that if you multiply the numerators together you will get the answer to the numerator of the third answer and if you add the first two numerators together it gets you the denominator of the third answer.  This then gives us the equation   to prove this I have created two problems to check the validity of the equation.

1. log39=c    log99=d    log279=

 = 2=      = 1=       = =

2. log381=c    log981=d    log2781=

= 4=      = 2=      = = =

This proves that if you use the numerator found in the first two problems of the set you can multiply them together to get the numerator of the third problem and if you add the first two numerators you can get the denominator of the third problem in each set.

Now the last thing to do is figure out if there is any sort of limitations of what a, b, and x can be.  To do this I will change the first set of logarithms , so there is an example of a positive, negative, fraction and decimal in place of the x and then the a.

log464=  log864=  log3264=

From this we already know that positive numbers will work.

      log4 -64=non-real    log8 -64=non-real    log32 -64=non-real

By making the x negative we get a non-real answer, so a negative x does not work.

Log-464=non-real    log-864=non-real    log-3264=non-real

By making the a negative we also get a non-real answer, therefore a putting a negative a or x we will get a non-real answer.

log4= -  log8= -  log32= -

By changing the x to a fraction it just changes the answer to a negative number.

log1/464= -     log1/864= -     log1/3264= -

By changing the a to a fraction it also just changes the answer to a negative number.  This also means that if both the a and x were fractions they would make a positive number.

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.