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$4^x - 4^{-x} \\ \mbox{I'm not sure what sort of answer you are looking for here. We can do this..} \\ 4^x = e^{\ln(4) x} \mbox{ and } 4^{-x} = e^{-\ln(4)x} \mbox{ so we have }\\ 4^x - 4^{-x} = e^{\ln(4) x} - e^{-\ln(4)x} = \\ 2\sinh(\ln(4)x)$