```Question 430444
<pre>
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:

{{{(5^(n+1)-5^(n-1))/(3*5^(3n))=a*5^(bn)}}}

{{{(5^(0+1)-5^(0-1))/(3*5^(3*0))=a*5^(b*0)}}}

{{{(5^1-5^(-1))/3*5^0=a*5^0}}}

{{{(5^1-1/5))/3*1=a*1}}}

{{{(5-1/5))/3=a}}}

{{{(25/5-1/5))/3=a}}}

{{{(24/5)/3=a}}}

{{{expr(24/5)*expr(1/3)=a}}}

{{{8/5=a}}}

---

Since it must hold for n=1, we substitute 1 for n, and {{{8/5}}} for a:

{{{(5^(1+1)-5^(1-1))/(3*5^(3*1))=expr(8/5)*5^(b*1)}}}

{{{(5^2-5^0)/(3*5^3)=expr(8/5)*5^b}}}

{{{(25-1)/(3*5^3)=8*(1/5)*5^b}}}

{{{24/(3*5^3)=8*5^(b-1)}}}

{{{8/(5^3)=8*5^(b-1)}}}

{{{1/5^3 = 5^(b-1)}}}

{{{5^(-3) = 5^(b-1)}}}

Equating exponents of 5

{{{-3 = b-1}}}

{{{-2 = b}}}

Edwin</pre>

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