Answer
$$y'=\frac{1}{\theta\ln 2}$$
Work Step by Step
$$y=\log_2(5\theta)$$
Recall Theorem 7: $$\frac{d}{d\theta}\log_au=\frac{1}{u\ln a}\frac{du}{d\theta}$$
Apply here to find $y'$: $$y'=(\log_2(5\theta))'=\frac{1}{5\theta\ln 2}(5\theta)'=\frac{5}{5\theta\ln 2}$$
$$y'=\frac{1}{\theta\ln 2}$$