Now I will check my answer:
=Log28 =Log48 =Log88
=Log8 =Log8 =Log8
Log2 Log4 Log8
=3 ∴ =1.5 ∴ =1 ∴
23=8 41.5=8 81=8
8=8 8=8 8=8
Later I decided I would apply the same method I used in the previous sequences towards different sequences and see if I would be successful with sequences to test the consistency of my pattern.
Ex. Log381, Log981, Log2781, Log8181......
The similarity between this pattern and the previous pattern is that the base is increasing with a consistent pattern. This time however the base of 3 is being raised by the term number.
6th = log(3^6)81 7th = log(3^7)81
=log24381 log72981
After continuing the pattern I can discover the general statement by converting the pattern from log form to exponential form.
3x=81, 9x=81, 27x=81, 81x=81, 243x=81, 729x=81
Now I will take this pattern one step further, by making the bases the same.
3nx=34, 3nx=34, 3nx=34, 3nx=34, 3nx=34, 3nx=34, 3nx=34
Through this pattern I concluded an expression of
nx=4
x=4/n
Now I am going to check the validity of my expression for log381 and log8181
x=4/n x=4/n
x=4/1 x=4/4
x=4 x=1
Now I will check my answer:
Log381 Log8181
=Log81 =Log81
Log3 Log81
=4 ∴ =1 ∴
34=81 811=81
81=81 81=81
To further investigate expressions for different sequences will considered also consider these last sequences..
Ex. log525, log2525, log12525, log62525, log312525, log1562525
Ex. logmmk, log(m^2)mk, log(m˄3)mk, log(m˄4)mk, log(m˄5)mk, log(m˄6)mk
Observing the previous sequences I will follow the same pattern and try to come up with and expression in the same form I have used for the pervious sequences.
5x=25, 25x=25, 125x=25, 625x=25, 3125x=25, 15625x=25
5nx=52, 5nx=52, 5nx=52, 5nx=52, 5nx=52, 5nx=52
Therefore the expression is nx=2 or x=2/n
Now I will check the validity of my expression with log2525, log62525
x=2/n x=2/n
x=2/2 x=2/4
x=1 x=0.5
Now I will check my answer
Log2525 log62525
=log25 =log25
Log25 log625
=1 ∴ =0.5 ∴
251=25 6250.5=25
25=25 25=25
m1(x)=mk, m2(x)=mk, m3(x)=mk, m4(x)=mk, m5(x)=mk, m6(x)=mk
mnx=mk, mnx=mk, mnx=mk, mnx=mk, mnx=mk, mnx=mk,
Therefore the expression is nx=k or x=k/n
Continuing my investigation I decided to look at other sequences however this time instead of
finding an expression that represents the sequence I will calculate the following sequences by giving my answer in p/q form.
Log464, log864, log3264
Log464 log864 log3264
4x=43 8x=82 32x=82
22x=22(3) 23x=22(3) 25x=22(3)
2x=6 3x=6 5x=6
x=6/2 x=6/3
x=3/1 x=2/1 x=6/5
log749, log4949, log34349
log749 log4949 log34349
7x=72 49x=49 343x=49
71x=72 72x=72 73x=72
1x=2 2x=2 3x=2
x=2/1 x=1/1 x=2/3
log(1/5)125, log(1/125)125, log(1/625)125
log(1/5)125 log(1/125)125 log(1/625)125
(1/5)x=125 (1/125)x=125 (1/625)x=125
5-1x=53 5-3x=53 5-4x=53
-1x=3 -3x=3 -4x=3
x= -3/1 x= -1/1 x= -3/4
log8512, log2512, log16512
log8512 log2512 log16512
8x=512 2x=512 16x=512
23x=29 21x=29 24x=29
3x=9 1x=9 4x=9
x=3/1 x=9/1 x=9/4
Looking at all these different sequences made me wonder if there was a relationship between the first and second answer that would help me obtain the third answer in that row. After observing different sequences it became apparent to me that there was a very obvious way I would be able to obtain the third answer from the previous two. For example, if I would first multiply the previous two answers and then divide it by the sum of the previous answers I will be able to conclude the third answer. To test my theory I will come up with two examples that fit the pattern above.
Log327, log927, log2727
For the third answer I use the method I had previously observed in the sequences above.
Log327 log927
3x=27 9x=27
31x=33 32x=33
1x=3 2x=3
x=3 x=3/2
Third answer = (3)(3/2)
3+(3/2)
= 1
Check validity for third answer:
Log2727
27x=27
33x=33
3x=3
X=1
Log216, log816, log1616
Again I will try to find the third answer using the method demonstrated above.
Log216 log816
2x=16 8x=16
21x=24 23x=24
1x=4 3x=4
x=4 x=4/3
Third answer = (4)(4/3)
4+(4/3)
= 1
Check validity for third answer:
Log1616
16x=16
24x=24
4x=4
X=1
If I look at the pattern I have created above and I let logax=c and logbx=d I could find a general statement that expressed logabx in the terms of c and d.
cd is equal to logab
c+d
Now that I have concluded a general statement to test the validity of my statement and to make sure it works with different values of c, d and x. I will be testing the validity by using other values of c, d and x.
logax = c, logbx = d, logabx = y
Log5125 = 3, log125125 = 1, log625125 = y
(cd)/(c+d) = y
(3 * 1)/(3+1) = y
¾=y
Now I am going to check my answer:
Log625125=y
Log125
Log625
¾ = y
After checking the validity of my general statement I reassured the fact that my general statement does work for multiple values. There are still some limitations to my general statement. For example my general statement only works for certain sequences that fit the same pattern of logax = c, logbx = d, logabx . This general statement does not work for all sequences for example it will work for the sequence of log28, log48, log88.
Now that I have come up with a successful general statement I would like to recall how I was initially able to come up with a general statement. When I look back at the first set of example I did with the longer sequences it helped me understand the way an expression could be formed in the format of p/q. It just gave me an overall understanding of how logs work and how you can convert logs into exponential which can be beneficial when solving for logs. Later when I moved on to my second set of example with the set of three terms in each sequences really helped me with finding the final general statement. My previous experience of forming expressions in the p/q form really helped me with these sequences. When I solved each term in all the sequences and put the answers in fraction form it became very obvious how I could use the first and second term answer to arrive at the third term answer. Seeing all the term answers side by side really helped me connect all the dots together. I was able to tell if I multiplied the two answers then divided them by the sum of the two terms I will be able to arrive at the answer.
