1. Statement of the problem. Find $\log_{10}(125.7)$ using logarithm tables. 2. Formula and important rules. For common logarithms we express a number as characteristic plus mantissa. The key relation is $$\log_{10} N = C + M$$ where $C$ is the characteristic (an integer determined by the order of magnitude) and $M$ is the mantissa (a positive fractional part found in the tables). When a number is written as $N = a \times 10^{n}$ with $1 \le a < 10$, then $$\log_{10} N = n + \log_{10} a$$ Use the mantissa table for $a$ and add the characteristic $n$. 3. Normalize the number. Write $125.7 = 1.257 \times 10^{2}$. So $$\log_{10}(125.7) = 2 + \log_{10}(1.257) $$ 4. Use the tables to find the mantissa. Look up the mantissa for $1.257$ in the log table by interpolating between nearby entries. From the table $1.250$ has mantissa $0.09691$ and $1.260$ has mantissa $0.10037$. Interpolate linearly because the table spacing is $0.01$. Compute the interpolation as $$M \approx 0.09691 + (1.257-1.250)\times \frac{0.10037-0.09691}{0.01}$$ Evaluate the arithmetic step by step. $1.257-1.250 = 0.007$ $0.10037-0.09691 = 0.00346$ Divide by $0.01$ to get $0.346$ and multiply by $0.007$ to get $0.002422$. So the mantissa is $M \approx 0.09691 + 0.002422 = 0.099332$. 5. Combine characteristic and mantissa. Thus $$\log_{10}(125.7) \approx 2 + 0.099332 = 2.099332$$ Round as desired, for example to five decimal places: $\log_{10}(125.7) \approx 2.09933$. 6. Final answer. Using logarithm tables and linear interpolation, $\log_{10}(125.7) \approx 2.09933$.