Is there a way to simplify a^log n, a = 1 / b

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The expression a^log[b]n can be simplified when a is defined as 1/b. The simplification leads to the result that (b^(-1))^log[b]n equals n^(-1). This transformation utilizes the property of exponents, specifically that (b^x)^y equals (b^y)^x. The discussion confirms that the simplification is valid and provides a clear mathematical explanation. Overall, the simplification effectively shows that a^log[b]n simplifies to 1/n.
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Is there a way to simplify a^logn, a = 1 / b

I don't know if this is possible, but I thought I would ask for another opinion if I'm overlooking something. Anyway, is there a way to simplify alogbn, a = 1 / b
 
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yes

\left(b^{-1}\right)^{log_{b} n} = \left(b^{log_{b} n}\right)^{-1} = n^{-1}


as

\left(b^{x}\right)^{y} = \left(b^{y}\right)^{x}
 
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aha! thanks!
 
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