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貌似f(x)=loga(x+b)/(x-b)吧? 令y=loga(x+b)/(x-b) 則有(x+b)/(x-b)=a^y x+b=xa^y-ba^y 得:x=b(1+a^y)/(a^y-1) 故反函數為y=b(1+a^x)/(a^x-1)
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