7. \( 2^{x^{2}-3} \cdot 5^{x^{2}-3}=0,01\left(10^{x-1}\right) \)

To solve the equation \( 2^{x^{2}-3} \cdot 5^{x^{2}-3}=0,01\left(10^{x-1}\right) \), we can simplify the right side first. We know that \( 0.01 = 10^{-2} \). Thus, we can rewrite the equation as: \[ 2^{x^{2}-3} \cdot 5^{x^{2}-3} = 10^{-2} \cdot 10^{x-1} \] This simplifies to: \[ 2^{x^{2}-3} \cdot 5^{x^{2}-3} = 10^{x-3} \] Since \( 10 = 2 \cdot 5 \), we can express \( 10^{x-3} \) in terms of its prime factors: \[ 10^{x-3} = (2 \cdot 5)^{x-3} = 2^{x-3} \cdot 5^{x-3} \] Now, equating the powers of 2 and 5 on both sides, we have: 1. For the base 2: \[ x^{2}-3 = x-3 \] 2. For the base 5: \[ x^{2}-3 = x-3 \] Both equations are identical, so we focus on solving one of them: \[ x^{2} - x = 0 \] Factoring gives us: \[ x(x-1) = 0 \] This leads to two solutions: \[ x = 0 \quad \text{or} \quad x = 1 \] Thus, the solutions to the original equation are \( x = 0 \) and \( x = 1 \).