1. **Problem Statement:** Expand $\log_e x$ in powers of $(x-1)$ and hence evaluate $\log_e 1.1$ correct up to 4 decimals. 2. **Formula and Explanation:** The Taylor series expansion of $\log_e x$ about $x=1$ is given by: $$\log_e x = \sum_{n=1}^\infty (-1)^{n+1} \frac{(x-1)^n}{n} = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots$$ This series converges for $|x-1| < 1$. 3. **Intermediate Work:** We want to find $\log_e 1.1$, so set $x=1.1$ and $h = x-1 = 0.1$. Calculate terms up to 4th power for accuracy: - First term: $h = 0.1$ - Second term: $-\frac{h^2}{2} = -\frac{(0.1)^2}{2} = -\frac{0.01}{2} = -0.005$ - Third term: $\frac{h^3}{3} = \frac{(0.1)^3}{3} = \frac{0.001}{3} \approx 0.0003333$ - Fourth term: $-\frac{h^4}{4} = -\frac{(0.1)^4}{4} = -\frac{0.0001}{4} = -0.000025$ Sum these terms: $$0.1 - 0.005 + 0.0003333 - 0.000025 = 0.0953083$$ 4. **Final Answer:** $$\log_e 1.1 \approx 0.0953$$ (correct up to 4 decimals) This matches the known value of $\log_e 1.1$ approximately.