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Chapter 26: Problem 757

If \(\log _{10} 3=.4771\) and \(\log _{10} 4=.6021\), find \(\log _{10} 12\).

Short Answer

Expert verified
Using the product rule for logarithms and the given values, we find \(\log _{10} 12 = \log _{10} 3 + \log _{10} 4 = .4771 + .6021 = 1.0792\).

Step by step solution

01

Express 12 in terms of 3 and 4

First, we'll write 12 as a product of 3 and 4, since we know the logarithms of these two numbers. We can write \(12 = 3 \times 4\).
02

Apply the product rule

Now, we'll apply the product rule for logarithms, which states that \(\log_b{(mn)} = \log_b{m} + \log_b{n}\). Applying this rule to our expression for 12, we get: \[ \log _{10} 12 = \log _{10} (3 \times 4) = \log _{10} 3 + \log _{10} 4 \]
03

Substitute the given values for logarithms

We are given the values of \(\log _{10} 3 = .4771\) and \(\log _{10} 4 = .6021\). We can substitute these values into our expression from Step 2: \[ \log _{10} 12 = .4771 + .6021 \]
04

Calculate the sum

Now we simply add the two values together: .4771 + .6021 = 1.0792
05

Write the final answer

Based on our calculations, the value of \(\log _{10} 12\) is 1.0792. Therefore: \[ \log _{10} 12 = 1.0792 \]

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