Using Lagrange's theorem, compute log sqrt {999}th root

Using Lagrange's theorem, compute $\log\sqrt[\sqrt{999}]{99}$ approximately.

Step 1: Reduce to two familiar quantities

\[ \log\sqrt[\sqrt{999}]{99} = \log 99^{1/\sqrt{999}} = \frac{\log 99}{\sqrt{999}}. \]

Both $99$ and $999$ sit just below numbers whose logarithm and square root are exact, so each factor is a mean value estimate over an interval of length $1$.

Step 2: The numerator, from $\log x$ on $[99,100]$

Since $\bigl(\log x\bigr)' = \dfrac{1}{x\ln10}$, Lagrange gives a $\xi\in(99,100)$ with

\[ \log 100-\log 99 = \frac{1}{\xi\ln10} \;\Longrightarrow\; \log 99 = 2-\frac{1}{\xi\ln10}. \]

Since $\xi$ is confined to a one-per-cent interval, taking $\xi \approx 100$: \[\log 99 \approx 2-\frac{1}{100\ln10} = 2-\frac{0.43429}{100} = 2-0.004343 = 1.995657.\]

Step 3: The denominator, from $\sqrt{x}$ on $[999,1000]$

With $\bigl(\sqrt{x}\bigr)' = \dfrac{1}{2\sqrt{x}}$, there is an $\eta\in(999,1000)$ with

\[ \sqrt{1000}-\sqrt{999} = \frac{1}{2\sqrt{\eta}} \;\Longrightarrow\; \sqrt{999} = \sqrt{1000}-\frac{1}{2\sqrt{\eta}}. \]

Taking $\eta \approx 1000$ and $\sqrt{1000} \approx 31.62278$: $\sqrt{999} \approx 31.62278-\frac{1}{63.24555} = 31.62278-0.01581 = 31.60697$.

Step 4: Divide

\[ \frac{\log 99}{\sqrt{999}} \approx \frac{1.995657}{31.60697} \approx 0.063140. \]

Step 5: Verification

\[\text{Exact: } \log 99 = 1.9956352,\quad \sqrt{999} = 31.6069613,\quad \text{quotient } = 0.0631391 \;\;\checkmark\]

\[\text{The estimates are high by } 2.2\cdot10^{-5} \text{ and } 9\cdot10^{-7} \text{ respectively} \;\;\checkmark\]

\[\text{Independent check: } \log 99 = 2+\log 0.99 = 2-0.0043648 = 1.9956352 \;\;\checkmark\]