If $\log_{10} 20 = 1.3010$ and $\log_{10} 30 = 1.4771$, then $\log_{10} (60000)$ is equal to

### Concept & Formula This question assesses your ability to factor a large number into the specific components provided in the prompt. By factoring $60000$ using $20$ and $30$, we can utilize the Product Rule of logarithms. Formula used: Product Rule: $\log(abc) = \log a + \log b + \log c$ ### Step-by-Step Solution Given the values: $$ \log_{10} 20 = 1.3010 $$ $$ \log_{10} 30 = 1.4771 $$ Step 1: Break down the number $60000$ into factors that include $20$, $30$, and powers of $10$. Notice that $20 \times 30 = 600$. To get $60000$, we need to multiply $600$ by $100$. $$ 60000 = 20 \times 30 \times 100 $$ Step 2: Apply the logarithm to both sides and use the Product Rule to expand the expression. $$ \log_{10} 60000 = \log_{10} (20 \times 30 \times 100) $$ $$ \log_{10} 60000 = \log_{10} 20 + \log_{10} 30 + \log_{10} 100 $$ Step 3: Substitute the known values and evaluate $\log_{10} 100$. (Since $100 = 10^2$, $\log_{10} 100 = 2$). $$ \log_{10} 60000 = 1.3010 + 1.4771 + 2 $$ Step 4: Add the numbers together. $$ \log_{10} 60000 = 2.7781 + 2 $$ $$ \log_{10} 60000 = 4.7781 $$ ### Exam Strategy & Shortcut **Direct Substitution via Scientific Notation:** You can also break $60000$ into $6 \times 10^4$. The log of $6 \times 10^4$ is simply $\log 6 + 4$. How to find $\log 6$? The problem gave $\log 20 = \log (2 \times 10) = \log 2 + 1 = 1.3010$, meaning $\log 2 = 0.3010$. Similarly, $\log 30 = \log 3 + 1 = 1.4771$, meaning $\log 3 = 0.4771$. $\log 6 = \log 2 + \log 3 = 0.3010 + 0.4771 = 0.7781$. Add the $4$ from the $10^4$, and you get $4.7781$. Both methods are equally valid and fast. ### Common Pitfall A common error is to over-factor the numbers into pure primes (like $2$ and $3$) without noticing that the given values ($\log 20$ and $\log 30$) can be used directly as larger "building blocks", saving calculation time. ### Final Answer **Therefore, the correct answer is 4.7781.**