If \(4^{x+1} = 64 \times 8^{x}\), find the value of \(x\).
-4
Express every term with base \(2\): \(4 = 2^2\), \(64 = 2^6\) and \(8 = 2^3\).
So \(4^{x+1} = (2^2)^{x+1} = 2^{2x+2}\) and \(64 \times 8^{x} = 2^{6} \times (2^3)^{x} = 2^{6+3x}\).
Equating the exponents of the equal powers of \(2\): \(2x + 2 = 6 + 3x\).
Solving, \(2 - 6 = 3x - 2x\), so \(x = -4\).
Hence, the value of \(x\) is \(-4\).
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