Exponential log equation for negative y

Taking logarithm base 2 on both sides: $\log_2 \left(2^{y^2 \log_3 5}\right) = \log_2 \left(5^{\log_2 3}\right)$ $y^2 \log_3 5 = \log_2 3 \cdot \log_2 5$

Since $\log_3 5 = \frac{\log_2 5}{\log_2 3}$: $y^2 \left(\frac{\log_2 5}{\log_2 3}\right) = \log_2 3 \cdot \log_2 5$ $y^2 = (\log_2 3)^2$

Since $y$ is negative: $y = -\log_2 3 = \log_2 \left(3^{-1}\right) = \log_2 \left(\frac{1}{3}\right)$