1. **State the problem:** Find the gradient of a line perpendicular to the line given by the equation $$8y + 4x = 5$$. 2. **Rewrite the equation in slope-intercept form:** We want to express the line as $$y = mx + c$$ where $$m$$ is the slope. Starting with: $$8y + 4x = 5$$ Subtract $$4x$$ from both sides: $$8y = -4x + 5$$ Divide both sides by 8: $$y = \frac{\cancel{8}y}{\cancel{8}} = \frac{-4x}{8} + \frac{5}{8}$$ Simplify the fraction: $$y = -\frac{1}{2}x + \frac{5}{8}$$ So, the slope $$m$$ of the original line is $$-\frac{1}{2}$$. 3. **Find the slope of the perpendicular line:** The slope of a line perpendicular to another is the negative reciprocal of the original slope. If original slope $$m = -\frac{1}{2}$$, then the perpendicular slope $$m_\perp = -\frac{1}{m} = -\frac{1}{-\frac{1}{2}} = 2$$. 4. **Final answer:** The gradient of the line perpendicular to $$8y + 4x = 5$$ is $$\boxed{2}$$.