If `log_(10)2 = 0.3010 and log_(10)7 = 0.8451,` then the value of `log_(10)2.8` is :

Options

A. A)0.4471
B. B)1.4471
C. C)2.4471
D. D)3.4471

Answer

To find the value of \( \log_{10} 2.8 \), we can use the properties of logarithms and the values given in the problem.

### Step-by-Step Solution:

1. **Express \( \log_{10} 2.8 \) in terms of known logarithms**:
\[ \log_{10} 2.8 = \log_{10} \left(\frac{28}{10}\right) \]
Using the property of logarithms that states \( \log_a \left(\frac{b}{c}\right) = \log_a b - \log_a c \):
\[ \log_{10} 2.8 = \log_{10} 28 - \log_{10} 10 \]

2. **Simplify \( \log_{10} 10 \)**:
Since \( \log_{10} 10 = 1 \), we have:
\[ \log_{10} 2.8 = \log_{10} 28 - 1 \]

3. **Express \( \log_{10} 28 \)**:
We can express 28 as \( 2^2 \times 7 \):
\[ \log_{10} 28 = \log_{10} (2^2 \times 7) \]
Using the property \( \log_a (b \times c) = \log_a b + \log_a c \):
\[ \log_{10} 28 = \log_{10} (2^2) + \log_{10} 7 \]

4. **Apply the power rule of logarithms**:
The power rule states \( \log_a (b^n) = n \cdot \log_a b \):
\[ \log_{10} (2^2) = 2 \cdot \log_{10} 2 \]
Therefore:
\[ \log_{10} 28 = 2 \cdot \log_{10} 2 + \log_{10} 7 \]

5. **Substitute the known values**:
We know that \( \log_{10} 2 = 0.3010 \) and \( \log_{10} 7 = 0.8451 \):
\[ \log_{10} 28 = 2 \cdot 0.3010 + 0.8451 \]

6. **Calculate \( \log_{10} 28 \)**:
\[ \log_{10} 28 = 0.6020 + 0.8451 = 1.4471 \]

7. **Substitute back to find \( \log_{10} 2.8 \)**:
Now, substituting back into the equation for \( \log_{10} 2.8 \):
\[ \log_{10} 2.8 = 1.4471 - 1 = 0.4471 \]

### Final Answer:
Thus, the value of \( \log_{10} 2.8 \) is:
\[ \boxed{0.4471} \]