### Instruction: Find $\log_{10} 40 +\log_{10} 25$.### Response: I recognize that this is a problem involving the properties of logarithms, specifically the product rule that says $\log_b (xy) = \log_b x + \log_b y$ for any positive base $b$ and any positive numbers $x$ and $y$.
Using this rule in reverse, I can rewrite the given expression as a single logarithm of a product: $\log_{10} 40 +\log_{10} 25 = \log_{10} (40 \cdot 25)$.
Now I need to simplify the product inside the logarithm.
I notice that both factors are multiples of 5, so I can factor out 5 from each: $40 \cdot 25 = 5 \cdot 8 \cdot 5 \cdot 5$.
This allows me to rewrite the product as a power of 5: $40 \cdot 25 = 5^3 \cdot 8$.
Next, I notice that 8 is a power of 2, so I can rewrite it as $2^3$: $40 \cdot 25 = 5^3 \cdot 2^3$.
Now I can use another property of logarithms, the power rule, that says $\log_b (x^y) = y \log_b x$ for any positive base $b$ and any positive number $x$ and any exponent $y$.
Applying this rule to both factors inside the logarithm, I get: $\log_{10} (40 \cdot 25) = \log_{10} (5^3 \cdot 2^3) = 3 \log_{10} 5 + 3 \log_{10} 2$.