1) \( \left(\frac{1}{4}\right)^{x-1}=16^{x+2} \)

To solve the equation \( \left(\frac{1}{4}\right)^{x-1}=16^{x+2} \), we first express both sides as powers of 2. We know that \( \frac{1}{4} = 2^{-2} \) and \( 16 = 2^4 \). Therefore, we can rewrite the equation as: \[ \left(2^{-2}\right)^{x-1} = \left(2^4\right)^{x+2} \] This simplifies to: \[ 2^{-2(x-1)} = 2^{4(x+2)} \] Now, we can set the exponents equal to each other since the bases are the same: \[ -2(x-1) = 4(x+2) \] Expanding both sides: \[ -2x + 2 = 4x + 8 \] Bringing all terms involving \( x \) to one side and constants to the other gives: \[ -2x - 4x = 8 - 2 \] Simplifying this, we have: \[ -6x = 6 \] Dividing both sides by -6 gives: \[ x = -1 \] Thus, the solution to the equation is \( x = -1 \).