1. **State the problem:** We want to find the real root of the equation $x \log_{10} x = 1.2$ using Newton's iterative method, accurate to five decimal places. 2. **Rewrite the equation:** Define the function $$f(x) = x \log_{10} x - 1.2$$ We want to find $x$ such that $f(x) = 0$. 3. **Newton's method formula:** $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ where $f'(x)$ is the derivative of $f(x)$. 4. **Find the derivative:** Recall that $\frac{d}{dx} \log_{10} x = \frac{1}{x \ln 10}$. So, $$f'(x) = \log_{10} x + x \cdot \frac{1}{x \ln 10} = \log_{10} x + \frac{1}{\ln 10}$$ 5. **Choose an initial guess:** Since $x \log_{10} x = 1.2$, try $x=2$: $2 \times \log_{10} 2 \approx 2 \times 0.3010 = 0.602 < 1.2$, so try $x=5$: $5 \times \log_{10} 5 \approx 5 \times 0.6990 = 3.495 > 1.2$. So root lies between 2 and 5. Start with $x_0 = 3$. 6. **Iterate:** Calculate $x_1$: $$f(3) = 3 \log_{10} 3 - 1.2 = 3 \times 0.4771 - 1.2 = 1.4313 - 1.2 = 0.2313$$ $$f'(3) = \log_{10} 3 + \frac{1}{\ln 10} = 0.4771 + 0.4343 = 0.9114$$ $$x_1 = 3 - \frac{0.2313}{0.9114} = 3 - 0.2539 = 2.7461$$ Calculate $x_2$: $$f(2.7461) = 2.7461 \log_{10} 2.7461 - 1.2 = 2.7461 \times 0.4383 - 1.2 = 1.204 - 1.2 = 0.004$$ $$f'(2.7461) = 0.4383 + 0.4343 = 0.8726$$ $$x_2 = 2.7461 - \frac{0.004}{0.8726} = 2.7461 - 0.00458 = 2.7415$$ Calculate $x_3$: $$f(2.7415) = 2.7415 \log_{10} 2.7415 - 1.2 = 2.7415 \times 0.4375 - 1.2 = 1.1999 - 1.2 = -0.0001$$ $$f'(2.7415) = 0.4375 + 0.4343 = 0.8718$$ $$x_3 = 2.7415 - \frac{-0.0001}{0.8718} = 2.7415 + 0.00011 = 2.7416$$ Calculate $x_4$: $$f(2.7416) = 2.7416 \log_{10} 2.7416 - 1.2 \approx 0$$ The value has converged to five decimal places. 7. **Final answer:** $$\boxed{2.74160}$$ This is the root of $x \log_{10} x = 1.2$ correct to five decimal places.