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Solve the equation 2^(3x)=8(4^(x-1/2)) for x.
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\[8\times 4^{x-\frac{1}{2}}=2^3\times (2^2)^{x-\frac{1}{2}}\] \[=2^3\times 2^{2x-1}\] \[=2^{2x-1+3}\] \[=2^{2x+2}\] is a start
Alright, then what? I don't really understand what to do in any equation if the exponent is a polynomial.
The bases on both sides are the same, so you can equate the powers. That is 3x = 2x+2 <- solve.
so x=2?
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