Compute approximately: 1° log_1011; 2° log_10998
Compute approximately: 1° $\log_{10}11$; 2° $\log_{10}998$.
Explanation of Method
Two logarithms, one close to its base point and one very close.
Step 1: The linearisation
With $f(x) = \log_{10}x$ we have $f'(x) = \dfrac{1}{x\ln10} = \dfrac{M}{x}$, where
\[ M = \log_{10}e = \frac{1}{\ln10} \approx 0.43429 \]
is the classical modulus of the decimal logarithm. Hence
\[ \log_{10}x \approx \log_{10}x_{0}+\frac{M}{x_{0}}\bigl(x-x_{0}\bigr). \tag{1} \]
The natural base points are the powers of ten, where the logarithm is an integer.
Step 2: Part 1°: $\log_{10}11$
Take $x_{0} = 10$, so $\log_{10}x_{0} = 1$ and $x-x_{0} = 1$: $\log_{10}11 \approx 1+\frac{0.43429}{10} = 1+0.043429 = 1.04343$.
Step 3: Part 2°: $\log_{10}998$
Take $x_{0} = 1000$, so $\log_{10}x_{0} = 3$ and $x-x_{0} = -2$: $\log_{10}998 \approx 3-\frac{2\times0.43429}{1000} = 3-0.00086859 = 2.99913$.
Step 4: Verification
\[\log_{10}11 = 1.0413927: \text{ the estimate is high by } 0.00204 \;\;\checkmark\]
\[\log_{10}998 = 2.9991305: \text{ the estimate is high by } 9\cdot10^{-7} \;\;\checkmark\]
\[\text{Ratio of relative steps: } \frac{1}{10} = 10\% \text{ against } \frac{2}{1000} = 0.2\%\]
Result
Final answer:
\[\boxed{\,\log_{10}11 \approx 1.0434, \qquad \log_{10}998 \approx 2.99913\,}\]
Remark
Remark. The two parts differ enormously in accuracy, and the table's last line says why: the error of a logarithm's linearisation is $-\tfrac{M}{2}\bigl(\tfrac{\Delta x}{x_{0}}\bigr)^{2}$, so it depends only on the relative step. For part 1° that is $-\tfrac{0.43429}{2}(0.1)^{2} = -0.00217$, matching the observed $0.00204$; for part 2° it is $-8.7\cdot10^{-7}$. A $10\%$ step is really too large for a first-order estimate — the $1.0434$ of part 1° is wrong in its third decimal.