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I am trying to solve the recurrence: $$ a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}} $$ but here is a problem for me. After few steps I have this: $$ a_n^2 = a_{n-1}\cdot a_{n-2} $$ and I don't now what to do further. I can solve a recurrence like that $$ a_{n+2} + a_{n+1} - a_n = 5 \cdot 2^n, $$ but I can't find any information about this case (when I have some degree or
square root in a recurrence).

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Hint: Let $b_n=\log a_n$ and solve a recurrence for $b_n$.

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$$ b_n=\log a_n , a_n^{b_n} = k ? $$ What do you mean? How it works? – Buga1234 Nov 28 '12 at 19:13
Where did you get $k$? If $b_n=\log a_n$ then $e^{b_n}=a_n$. – Thomas Andrews Nov 28 '12 at 19:18
:) Now I've the same question. Where did you get that $ e $ ? – Buga1234 Nov 28 '12 at 19:24
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You don't need to use natural logarithms, you can use any logarithm base you want with this problem. If you define $b_n=\log_{10} a_n$ then you know that $10^{b_n}=a_n$. – Thomas Andrews Nov 28 '12 at 19:58
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yeap... it must be $ 2b_n = b_{n-1} + b_{n-2} $ – Buga1234 Nov 28 '12 at 20:24
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