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So is it this: 64 = 4[ (1/2)^(3x - 1) ]. First I would divide by 4: 16 = (1/2)^(3x - 1). Then you could recognize that 1/2 is the same as 2^(-1) so we have this:

16 = 2^(1 - 3x). Then you might notice that 2^4 is 16, if not then you could take log [base 2] of both sides.

2^4 = 2^(1-3x), So the exponents must equal each other: 4 = 1-3x. so we have 3 = -3x and then x = -1. Plug back in and it checks out

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9y ago
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Q: What is the solution to 64 equals 4 times one half with an exponent of 3x-1?
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