1) \( (0,1)^{2 x-3}=10 \);

To solve the equation \( (0,1)^{2x-3} = 10 \), we first need to clarify the expression \( (0,1)^{2x-3} \). The notation \( (0,1) \) typically refers to the open interval between 0 and 1, which is not a valid base for exponentiation in this context. However, if we interpret \( (0,1) \) as a base \( b \) where \( 0 < b < 1 \), we can proceed with the solution. Let's assume \( b \) is a constant in the interval \( (0,1) \). The equation can be rewritten as: \[ b^{2x-3} = 10 \] Since \( b \) is between 0 and 1, \( b^{2x-3} \) will always be positive and less than or equal to 1 for any real number \( x \). Therefore, it cannot equal 10, which is greater than 1. Thus, there are no solutions to the equation \( (0,1)^{2x-3} = 10 \) under the assumption that \( (0,1) \) represents a base in the interval \( (0,1) \). If you meant something different by \( (0,1) \), please clarify, and I would be happy to help further!