SOLUTION: FInd the value of a and b if (5^(n+1)-5^(n-1))/(3x5^(3n))=ax5^(bn)

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Question 430444: FInd the value of a and b if
(5^(n+1)-5^(n-1))/(3x5^(3n))=ax5^(bn)
Found 2 solutions by robertb, Edwin McCravy:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
%285%5E%28n%2B1%29+-+5%5E%28n-1%29%29%2F%283%2A5%5E%283n%29%29+=+a%2A5%5E%28bn%29
<==> %2825%2A5%5E%28n-1%29+-+5%5E%28n-1%29%29%2F%283%2A5%5E%283n%29%29=+a%2A5%5E%28bn%29
<==> %2824%2A5%5E%28n-1%29%29%2F%283%2A5%5E%283n%29%29=+a%2A5%5E%28bn%29
<==> 8%2A5%5E%28-2n-1%29+=+a%2A5%5E%28bn%29
==> %288%2F5%29%2A5%5E%28-2n%29+=+a%2A5%5E%28bn%29
==> a+=+8%2F5 and b = -2, by direct comparison.


Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
Another approach:
There are three variables and one equation.
Therefore if no other restrictions apply, and
if irrational solutions are allowed, then there
are infinitely many solutions for instance.  
n=-2.5, a=.2139968976, b=-2.5
n = -3, a=1, b=-2.097343225

Therefore it is possible only to do one of these
three things:

1. Solve for a in terms on n and b
2. Solve for b in terms of n and a
3. Solve for n in terms of a and b
4. Assume the equation is an identity and holds for all values of n

To avoid infinitely many solutions, I will do only the last.  

Since it must hold for all n, if must hold for n=0, 
so we substitute 0 for n:
  
%285%5E%28n%2B1%29-5%5E%28n-1%29%29%2F%283%2A5%5E%283n%29%29=a%2A5%5E%28bn%29

%285%5E%280%2B1%29-5%5E%280-1%29%29%2F%283%2A5%5E%283%2A0%29%29=a%2A5%5E%28b%2A0%29

%285%5E1-5%5E%28-1%29%29%2F3%2A5%5E0=a%2A5%5E0

%285%5E1-1%2F5%29%29%2F3%2A1=a%2A1

%285-1%2F5%29%29%2F3=a

%2825%2F5-1%2F5%29%29%2F3=a

%2824%2F5%29%2F3=a

expr%2824%2F5%29%2Aexpr%281%2F3%29=a

8%2F5=a

---

Since it must hold for n=1, we substitute 1 for n, and 8%2F5 for a:
  


%285%5E2-5%5E0%29%2F%283%2A5%5E3%29=expr%288%2F5%29%2A5%5Eb

%2825-1%29%2F%283%2A5%5E3%29=8%2A%281%2F5%29%2A5%5Eb

24%2F%283%2A5%5E3%29=8%2A5%5E%28b-1%29

8%2F%285%5E3%29=8%2A5%5E%28b-1%29

1%2F5%5E3+=+5%5E%28b-1%29

5%5E%28-3%29+=+5%5E%28b-1%29

Equating exponents of 5

-3+=+b-1

-2+=+b 

Edwin