Question

$(2^{x-1}*4^{x+1})/8^{x-1})=64$

Answer 1

Step-by-step explanation:

$\frac{ {2}^{(x - 1) } \times {4}^{(x + 1)} }{ {8}^{(x - 1)} } = 64 \\ or. \: \: \frac{ {2}^{(x - 1)} \times {2}^{(2x + 2)} }{ {2}^{(3x - 3)} } = 64 \\ or. \: \: {2}^{(x - 1) + (2x + 2) - (3x - 3)} = {2}^{6} \\ or. \: \: (x - 1 + 2x + 2 - 3x + 3) = 6 \\ or. \: \: 3x - 3x + 4 = 6 \\ or. \: \: 4 = 6$

This is a wrong Equation...