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Solve 64^(2-x)=4^(4x)
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\[\large64=4^4\]so we have\[64^{2-x}=4^{4x}\]\[(4^4)^{2-x}=4^{4x}\]\[4^{8-4x}=4^{4x}\]do you think you can solve it now?
8-4x= 4x add 4x to each sides
(4^4)^(2-x) =4^(4x) 4^(8-4x) =4^(4x) 8-4x=4x 8=8x x=8/8 x=1
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