1. Problem statement: Find the equation of the normal to the curve $\ln(2x^2 - 7)$ at the point where the curve crosses the positive x-axis. 2. Find the x-intercept by setting the curve equal to 0 because the x-axis is $y=0$. $$\ln(2x^2 - 7) = 0$$ Exponentiate both sides to remove the logarithm. $$2x^2 - 7 = e^0 = 1$$ Solve for $x$ by isolating $x^2$. $$2x^2 = 8$$ Show the division step and simplify the fraction using cancellation. $$x^2 = \frac{8}{2}$$ $$x^2 = \frac{\cancel{8}}{\cancel{2}} = 4$$ Take square roots to find the x-intercepts. $$x = \pm 2$$ The positive x-intercept is $x = 2$, so the point of interest is $$(2,0)$$. 3. Compute the derivative to get the slope of the tangent using the chain rule; for $y=\ln(u)$ we have $y'=u'/u$. $$y' = \frac{d}{dx}\ln(2x^2 - 7) = \frac{4x}{2x^2 - 7}$$ 4. Evaluate the derivative at $x=2$ to obtain the tangent slope. $$y'(2) = \frac{4\cdot 2}{2\cdot 2^2 - 7} = \frac{8}{8-7} = \frac{8}{1}$$ Show the simplification using cancellation. $$\frac{8}{1} = \frac{\cancel{8}}{\cancel{1}} = 8$$ So the tangent slope at $(2,0)$ is $8$. 5. The normal line has slope equal to the negative reciprocal of the tangent slope. $$m_{n} = -\frac{1}{8}$$ 6. Use the point-slope form of a line for the normal through the point $$(2,0)$$. $$y - 0 = -\frac{1}{8}(x - 2)$$ Clear denominators by multiplying both sides by 8. $$8y = -(x - 2)$$ Expand the right-hand side. $$8y = -x + 2$$ Rearrange to integer-coefficient form $ax + by + c = 0$. $$x + 8y - 2 = 0$$ 7. Final answer: $x + 8y - 2 = 0$.