If the n -th significant digit of an approximate value x

If the $n$-th significant digit of an approximate value $x$ is in error by at most half a unit in that place, find a bound for the error of the mantissa of $\log_{10}x$.

Step 1: What "half a unit in the last significant place" means

Write $x$ in the form $x = d_{1}.d_{2}\ldots d_{n}\times 10^{k}$ with leading digit $1\le d_{1}\le 9$, so that $10^{k}\le x\lt 10^{k+1}$. The $n$-th significant digit sits in the place $10^{\,k-n+1}$, and the hypothesis is

\[ \bigl|\Delta x\bigr| \le \tfrac12\cdot 10^{\,k-n+1}. \tag{1} \]

Step 2: Convert (1) into a bound on the relative error

Since $x\ge 10^{k}$, dividing (1) by $x$ gives

\[ \frac{\bigl|\Delta x\bigr|}{x} \le \frac{\tfrac12\cdot 10^{\,k-n+1}}{10^{k}} = \frac{1}{2}\cdot 10^{\,1-n} = \frac{1}{2\cdot 10^{\,n-1}}. \tag{2} \]

This is the crucial step: the absolute error depends on the exponent $k$, but the relative error does not — it is controlled by the number of significant digits alone.

Step 3: The differential of the decimal logarithm

The mantissa is the fractional part of $\log_{10}x$; changing $x$ slightly changes the mantissa by the same amount as it changes $\log_{10}x$ itself. With $\bigl(\log_{10}x\bigr)' = \dfrac{1}{x\ln 10}$:

\[ \bigl|\Delta\bigl(\log_{10}x\bigr)\bigr| \approx \frac{\bigl|\Delta x\bigr|}{x\ln 10} = M\cdot\frac{\bigl|\Delta x\bigr|}{x}, \qquad M = \log_{10}e = \frac{1}{\ln 10} \approx 0.43429. \]

Step 4: Combine with (2)

\[ \bigl|\Delta\bigl(\log_{10}x\bigr)\bigr| \le \frac{M}{2\cdot 10^{\,n-1}} = \frac{1}{2\ln 10\cdot 10^{\,n-1}} \approx \frac{0.2171}{10^{\,n-1}}. \]

Step 5: When the bound is attained

Step 2 used the weakest inequality $x\ge 10^{k}$, i.e. the worst case $d_{1} = 1$. For a general leading digit the sharper bound is

\[ \bigl|\Delta\bigl(\log_{10}x\bigr)\bigr| \lesssim \frac{M}{2\,d_{1}\cdot 10^{\,n-1}}, \]

so a number beginning with $9$ has its logarithm about nine times better determined than one beginning with $1$.

Step 6: Verification

\[n = 4,\ x = 1.000:\quad \bigl|\Delta x\bigr| \le 0.0005, \qquad \log_{10}(1.0005)-\log_{10}(1) = 0.000217\]

\[\text{Bound: } \frac{0.2171}{10^{3}} = 0.000217 \;\;\checkmark \quad\text{(attained, as } d_{1}=1\text{)}\]

\[n = 4,\ x = 9.000:\quad \log_{10}(9.0005)-\log_{10}(9) = 0.0000241 \approx \frac{0.2171}{9\cdot 10^{3}} \;\;\checkmark\]