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I need some hint towards calculating $e^{iAx}Be^{-iAx}$ Given that $[A,B]=B$

I have been manipulating this (by expansion) for quite a while, and am wondering if there is any clever way to approach this problem. In the expansion, just the first order terms seem to suggest a solution like $e^{i Bx}$ but higher order terms mess it up. Besides the brute force, the only thing I have seen here is that if $X=e^{iAx}Be^{-iAx}$ then $[A,X] = X$. So X might be B times some factor, though that is not a guarantee. Any hints would be welcome.

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    hmm... going over wikipeidia, there's another method.. to use $ e^{A}Be^{-A}=B+[A,B]+\frac{1}{2!}[A,[A,B]]+\frac{1}{3!}[A,[A,[A,B]]]+\cdots $2011-10-18

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Hint: what is $\frac{d}{dx} e^{iAx}Be^{-iAx}$?