Using $\log x+\log y=\log xy,$ we get that $\log_{10} 40+\log_{10} 25=\log_{10}(40\cdot 25)=\log 1000.$ That means we want $x$ where $10^x=1000,$ which means $x=3.$ Therefore, $\log_{10} 40+\log_{10} 25=\boxed{3}.$
Find $\log_{10} 40 +\log_{10} 25$.
{'answer': '3', 'subject': 'Intermediate Algebra', 'level': 1, 'unique_id': 'train/intermediate_algebra/339.json'}