Therefore the expression is: Log 2^n 8 = 3 / n
Sequence 2:
Explanation for the next two terms: # 5 and 6 terms are continuation of the series and they follow a pattern. The Argument (A) remains the same throughout the series – 81. The pattern with the bases was 3^n hence the 5 Th and 6 Th term bases being 35 and 36 which is 243 and 729 which were put as bases for the respective terms.
To find the nth term of this sequence in form of p/q: Log 381, Log 9 81, Log 27 81, Log 81 81, Log 243 81, Log 729 81
The bases for the consecutive terms show pattern and can be written as 31 for the first term, 32 for the second term as so on till 36 for the 6th term. Hence bases can be generally denoted by 3n where n is the number of term (similar to the previous series).
Hence the logarithm equation for the nth term is: Log 3^n 81
Let Log 3^n 81 = y, y= nth term in p/q form.
Then according to logarithm law:
3ny= 81 → 3ny= 34
ny= 4(according to exponents law) and y= 4/n
Therefore the expression is: Log 3^n 81 = 4 / n
Sequence 3:
For the 5th and the 6th term only the base changes as the Argument remains the same with the previous four terms in the series. The pattern followed by the bases is the same we have figured so far and in this case the general base formula is 5n. Hence for the 5th and the 6th terms have the bases 55 and 56(3125 and 15625) respectively and 25 as the same argument.
To find the nth term of this sequence in form of p/q:  Log 5 25, Log 25 25, Log 125 25, Log 625 25, Log 3125 25, Log 15625 25
The argument remaining the same i.e. 25, the bases in this sequence can be written generally in the form 5^n ( 51 for the first term, 52 for the second up till 56 for the sixth)
Hence the logarithm equation for the nth term is: Log 5^n 24
Let Log 5^n 25 = y, y= nth term in p/q form.
Then according to logarithm law:
5ny= 25 → 5ny= 52
ny= 2(according to exponents law) and y= 2/n
Therefore the expression is: Log 5^n 25 = 2 / n
Sequence 4:
The 5th and 6th term are in sequence of the previous four with Argument mk repeated and bases show pattern with base m and consecutive exponent. The generalised base in this case is mn and hence the 5th and 6th term have the bases m5 and m6 respectively.
For the nth term of this sequence let
Log m^n mk = y, y= nth term in p/q form.
Then according to logarithm law:
mny= mk
ny= k(according to exponents law) and y= k/n
Therefore the expression is: Log m^n mk = k / n
For Justification: (Using Algebra):
Part 2: The Task:
In the following calculations the basic law of logarithm is applied (based on the argument, bases and exponents) :  logarithm law which states that if logab= c then ac=b
Answers as given in p/q form where p and q belong to rational numbers.
Sequence 1 : Log464, Log864, Log3264
1)Let Log464 = x; then 4x= 64 => 4x =43
Hence x = 3/1 (according to the exponents law)
2)Similarly, Log864 = x2; 8x=64 => 8x=82
Hence x2= 2/1 (in p/q form)
3)Log3264 =x3 ; then 32x=64 =>25x=26
Hence 5x=6 so x3=6/5
Sequence 2: Log749, Log4949, Log34349
1)Let Log749=p; then 7p=49 => 7p=72
Therefore p=2/1
2)Log4949 =r: as base =argument (49 raised to 1 would be 49 itself)
Hence answer=r= 1/1
3)Log34349= q; 343q=49=>73q=72
3q=2 and q=2/3
Sequence 3: Log1/5125, Log1/125125, Log1/625125
1)Let Log1/5125= a; (1/5)a= 125 5a =53 ((1/a)= a1))
Hence a= 3/1
2) Log1/125125=b; 125b=1251
Hence b=1/1
3) Log1/625125= c; (1/625)c= 12554c=53
Hence c= 3/4
Sequence 4: Log8512, Log2512, Log16512
1) Let Log8512 =d; 8d=512(2^9) 23d=29
Hence d= 9/3 => d=3/1
2)Log2512 =e; 2e= 512 2e=29
Hence e=9/1
3)Log16512=f; 16f=512 24f=29
Hence f=9/4
General Trend : To obtain the third answer in each row the bases of the first two terms get multiplied to obtain the base of the third while the Argument remains the same throughout.
For example in the fourth sequence the base of first term 8 gets multiplied with base of the second term (2) to get the base of the third term 8 X 2 = 16;while the argument512, remains the same.
Two more examples that fit the pattern above:
1) Log4128, Log16128, Log64128 (4 x 16 = 64)
Base of 1st term
2) Log28, Log48, Log88 (2 X 4= 8)
Part 3: The Task:

Let logax =c ;
Then according to the log law;
= c
Then log a =
Let logbx =d ;
Then;
= d
Hence log b =
Now taking the other log equation (of which the general term has to be found) and substituting the above terms to mutually cancel out the log terms and obtain the general term in pure variable form which express log ab x.
Log ab x ; which can be written as 
(according to the log law)
Now using the log identity – Log b (m•n) = Log b m + Log b n; log ab= log a + log b (base 10 or k)
Hence, log abx=
Substituting log a and log b from the 1 and 2 equations derived in the previous part:
=Log abx=
= Log abx=
(TAKING L.C.M AND MULTIPLYING)
= Log abx=
(taking log x common)
= Log abx= log
X
Hence, Log abx=
is the general statement that expresses logabx in terms of c and d.
*Testing the validity of the general statement:
1 2
Scope/limitations of a , b , and x in this equation:
 In logabx= cd/c+d; x cannot be a negative number as no positive number put to an exponent can give a negative result. Hence it is undefined.
 if x is zero; equation is undefinedlogab0=cd/c+d ; as no power raised to a number would give zero as a result.
if x=1; then logab 1= 0 (as anything raised to zero is one)
Conclusion:
To conclude the portfolio; the concept of logarithm laws and bases has been successfully applied to complete the tasks and proceed with the equations and to find the general statement.
Overall concept of logarithms and relating it with exponents has been carried out well.