log23 / log23= 3log2 / 3log2 = 3/3 = 1
log23 / log24= 3log2 / 4log2 = 3/4
log23 / log25= 3log2 / 5log2 = 3/5
After evaluating the logarithms, you can see that the denominator of each value is the same as the term number. Therefore the conclusion that the value of the sequence can be calculated by using the expression 3/n where n is the term number. Using this expression, the value of the 6th and 7th terms may be found.
Value of 6th term Value of 7th term
3/6 3/7
Multiply the ratios by log2 and rewrite the logarithm using Change-the-Base.
3log2 / 6log2 3log2 / 7log2 multiply log2 to both sides
log23/ log26 log23 / log27 change multiplier to exponent
log8 / log64 log8 / log128 expand exponential
log648 log1288 change the base to rewrite
logarithm
In the second, third, and fourth sequences, the setup of the terms is very similar to that of the first sequence where the base of the logarithm changes but not the . Because of that, the sequences are also both geometric sequences. Finding the common ratio can further prove this theory to be correct.
Second Sequence…
log381 , log981 , log2781 , log8181 , …
9/3 = 3 , 27/9 = 3 , 81/27 = 3 the common ratio of the sequence is 3
Third Sequence…
log525 , log2525 , log12525 , log62525 , …
25/5 = 5 , 125/25 = 5 , 625/125 = 5 the common ratio of the sequence is 5
Fourth Sequence…
logmmk , logm^2mk , logm^3mk , logm^4mk , …
m2/m = m , m3/m2 = m , m4/m3 = m the common ratio of the sequence is m
Now you can evaluate the terms in the form p/q.
Second Sequence
As the sequence shows, the general expression used to obtain the answer is 4/n.
Value of 5th term Value of 6th term
4/5 4/6
4log3 / 5log3 4log3 / 6log3 multiply log3 to both sides
log34 / log35 log34 / log36 change multiplier into exponent
log81 / log243 log81 / log729 expand exponential
log24381 log72981 change the base to rewrite logarithm
Third Sequence
As the sequence shows, the general expression used to obtain the answer is 2/n.
Value of 5th term Value of 6th term
2/5 2/6
2log5 / 5log5 2log5 / 6log5 multiply log3 to both sides
log52 / log55 log52 / log56 change multiplier into exponent
log25 / log3125 log25 / log15625 expand exponential
log312525 log1562525 change the base to rewrite logarithm
Fourth Sequence
To evaluate the 5th and 6th terms in the sequence, only the n in the formula logmnmk needs to be replaced because the sequence’s logarithms are already in the exact same form.
So due to the information provided by the four sequences, we can come to the conclusion that the formula of logmnmk can be replaced with the expression k/n where k is a constant.
Below is a graph showing the values of the three sequences
Part 2
Given the following sequences, there is a relationship between the 3 terms in each sequence.
log464 , log864 , log3264
log749 , log4949 , log34349
log1/5125 , log1/125125 , log1/625125
log8512 , log2512 , log16512
When closely examined, you will notice that the first logarithm’s base multiplied by the second logarithm’s base results in the third logarithm’s base.
First Sequence
(4)(8) = 32
Second Sequence
(7)(49) = 343
Third Sequence
(1/5)(1/125) = 1/625
Fourth Sequence
(8)(2) = 16
Values found using the base numbers of the first and second logarithms.
Different values can be used for the nth terms in a sequence. Below will be two examples of sequences that follow this same pattern.
Sequence #1
log6216 , log3216 , log18216
(6)(3) = 18
Sequence #2
log232 , log532 , log1032
(2)(5) = 10
Multiply the bases of the first and second logarithms to find the base of the third
logarithms.
Part 3
Given that logax = c and logbx = d, find a general statement that expresses logabx in terms of c and d.
logax = c , logbx = d
(ac)d = xd , (bd)c = xc rewrite the logarithms as exponential form
acd = xd , bcd = xc multiply the exponents
(ab)cd = x(c+d) multiply the two equations together
(cd)logab = (c+d)logx rewrite exponentials as logarithms
(cd)logab/c+d = logx divide c+d by both sides to get logx by itself
cd/c+d = logx / logab divide logab by both sides
cd/c+d = logabx rewrite the right half of the equation as a logarithm thus producing the general statement that expresses logabx in terms of c and d
As the work shows, cd/c+d is the general statement needed to be found. We can test the validity of this statement by using various values for a, b, and x. For example…
a = 2 , b = 4 , x = 8
logax = log28
logbx = log48
logabx = log88
log21(23) , log22(23) , log23(23)
3 , 3/2 , 1
c = 3 , d = 3/2
cd/c+d
= (3)(3/2)/(3+3/2)
=1 the expression cd/c+d is valid because the result is the same as the result of logabx, which was shown 2 steps before. (1=1)
Even now when it has been proven that the expression cd/c+d is equivalent to the expression logabx, there are still some things that need to be set in order. There are a series of limitations on these expressions that must be shown so as not to have any values that make the expression false.
Limitations
The first limitation on cd/c+d = logabx is that ab>0, the reason being because a logarithm cannot be negative. For example if…
a = -1 b = 5 x = 125
logabx
= log(-1)(5)125 substitute the values in for a, b, and x
= log-5125 combine
= log125/log-5 by using change-of-base, you can clearly see the negative logarithm which does not exist
Another limitation of cd/c+d = logabx is that ab = 1. This cannot be because log1 is equal to zero. If…
a = 1 b = 1 x = 125
logabx
= log1125 substitute in the values for a, b, and x
= log125 / log1 change-of-base to be able to see the work of the equation easily
=log125/ 0 solve log1, which equals to 0.
Yet another limitation of cd/c+d = logabx is that ab = 0. This cannot be because log0 is not possible. If…
a = 0 b = 5 c = 125
logabx
= log(0)(5)125 substitute in the values for a, b, and x
= log0125 combine
= log125 / log0 change-of-base to be able to see the work of the equation easily.
Not possible since log0 does not exist.
The next limitation of cd/c+d = logabx is that x>0. It is a limitation for the same reasons as the first limitation: a logarithm cannot be negative.
The last limitation on cd/c+d = logabx is that x = 0. This limitation is also a limitation for the same reasons as the second limitation: a logarithm cannot equal zero.
Therefore, from the information provided, you can conclude that the general statement works only with positive numbers.
Conclusion
In conclusion, from the information in this investigation, we can say that the value of the logarithmic sequences in the form of p/q can be reached through the equation k/n. Regularly, the value of logarithmic sequences can be reached through the equation logabx. We also have determined what the general statement was to the expression logabx. The general statement is cd/c+d which has a few limitations that are ab>0, ab = 1, ab = 0, x>0, and x = 0.