Compute ln 3 approximately, then use the result

Compute $\ln 3$ approximately, then use the result to compute $3^{\sqrt{4.3}}$.

Step 1: Linearise the logarithm at $x = e$

The nearest point at which $\ln x$ is known exactly is $x = e$, where $\ln e = 1$ and $\bigl(\ln x\bigr)' = \dfrac1x$ equals $\dfrac1e$: \[\ln x \approx 1+\frac{x-e}{e} = \frac{x}{e}. \tag{1}\]

The linearisation of the logarithm at $e$ is therefore just division by $e$ — a memorable form.

Step 2: Evaluate at $x = 3$

\[ \ln 3 \approx \frac{3}{e} = \frac{3}{2.718282} \approx 1.10364. \tag{2} \]

The true value is $1.09861$, so this is high by $0.00503$ — about $0.46\%$. The error is respectable given that $3$ is not especially close to $e$.

Step 3: The exponent, from Problem 260

\[ \sqrt{4.3} = \sqrt{2^{2}+0.3} \approx 2+\frac{0.3}{4} = 2.075. \tag{3} \]

Step 4: Split off the integer part

Writing $2.075 = 2+0.075$ isolates a small exponent: \[3^{2.075} = 3^{2}\cdot3^{0.075} = 9\cdot3^{0.075}. \tag{4}\]

Step 5: Linearise $3^{\alpha}$ at $\alpha = 0$

With $g(\alpha) = 3^{\alpha} = e^{\alpha\ln3}$ we get $g(0) = 1$ and $g'(0) = \ln3$, so

\[ 3^{\alpha} \approx 1+\alpha\ln3, \qquad 3^{0.075} \approx 1+0.075\times1.10364 = 1.08277. \tag{5} \]

Step 6: Combine

\[ 3^{\sqrt{4.3}} \approx 9\times1.0828 = 9.745. \]

Step 7: Verification

\[\text{Exact: } \sqrt{4.3} = 2.0736441,\ \ln3 = 1.0986123,\ \text{product } = 2.2781309\]

\[3^{\sqrt{4.3}} = e^{2.2781309} = 9.7584242\]

\[\text{Estimate } 9.745 \text{ is low by } 0.0132, \text{ i.e. } 0.135\% \;\;\checkmark\]