Question 155464
Let's look at the first part, a:
a) {{{Log[3](81)}}},{{{Log[9](81)}}},{{{Log[27](81)}}},{{{Log[81](81)}}},...{{{Log[n](81)}}}
As you can see, the base (3 in this case) is raised to a power that is equal to the term number.
In the 1st term, the base, 3, is raised the the first power or {{{3^1 = 3}}}
In the second term, the base is raised to the second power or {{{3^2 = 9}}} and so on to the nth term, the power will be raised to the nth power or {{{3^n}}}
Now you can work out the values of the first few terms by remembering the definition of a logarithm of a number.
"The logarithm of a number is the power to which the base of the logarithm must be raised to equal the number"
So, for the first term (n=1), we have:
{{{Log[3](81) = x}}} or...
{{{3^x = 81}}} but we can express 81 as multiples of 3...{{{81 = 3^4}}}, so we;ll substitute:
{{{3^x = 3^4}}} therefore, {{{x = 4}}} and so...
{{{Log[3](81) = 4/1}}} for n = 1
Similarly for the second term (n=2), we have:
{{{Log[3^2](81) = x}}} and...
{{{3^2x = 81}}} or...
{{{3^2x = 3^4}}} and so...
{{{2x = 4}}} and {{{x = 4/2}}} so we get:
{{{Log[9](81) = 4/2}}} for n = 2
Let's look at the nth term:
{{{Log[3^n](81) = x}}} or
{{{3^nx = 3^4}}} so that:
{{{nx = 4}}} and 
{{{x = 4/n}}} so you can write:
{{{Log[3^n](81) = 4/n}}} for the nth term. Where n is an integer =1, 2, 3, ...n
So if you apply the same logic to the rest of the problems, you should get the required answers.
Let's look at c)
As in the previous problems, the power of the base is equal to the number of the term, n.
{{{Log[m^n](m^k) = x}}} or...
{{{m^nx = m^k}}} so that...
{{{nx = k}}} and 
{{{x = k/n}}} so...
{{{Log[m^n](m^k) = k/n}}}
If you are still having difficulties with these, please repost.