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If 2^x=3, whats 4^-x=?
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2^x = 3 4^-x = 2^(-2x) = 3^-2 = 1/9
wait how does 4^-x= 2^(-2x)? and where do the 2 and x when you get to 3^-2?
4^-x = 2^-2x because 4 = 2^2 so if you manipulate, 4^-x = 2^2(-x) correct? so 4^-x = 2^-2x now i got 3^-2 from the previous equation 2^x = 3 so when 2^-2x, for both sides to be equal, you must also raise the other side by -2 so i got 3^-2. Did that answer your question?
yeah thanks
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