How do you solve 10^(x-3)>=52?

$x \ge \log 52 + 3$
First, let ${10}^{x - 3} = 52$ From then, we can add $\ge$ later on. Solving this equation, ${\log}_{10} 52 = x - 3$ or $\log 52 = x - 3$. $\therefore x = \log 52 + 3$. Adding back the $\ge$, $x \ge \log 52 + 3$