1. **State the problem:** We are given the function $y=\ln(x^2+4k)$ and need to find the value of $k$ such that the slope of the tangent line at $x=1$ is 15. 2. **Recall the formula for the derivative:** The slope of the tangent line to $y=\ln(u)$ is given by $\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}$. 3. **Apply the chain rule:** Here, $u = x^2 + 4k$, so $\frac{du}{dx} = 2x$ because $k$ is a constant. 4. **Write the derivative:** $$\frac{dy}{dx} = \frac{1}{x^2 + 4k} \cdot 2x = \frac{2x}{x^2 + 4k}$$ 5. **Evaluate the derivative at $x=1$:** $$\left. \frac{dy}{dx} \right|_{x=1} = \frac{2 \cdot 1}{1^2 + 4k} = \frac{2}{1 + 4k}$$ 6. **Set the slope equal to 15 and solve for $k$:** $$\frac{2}{1 + 4k} = 15$$ 7. **Cross multiply:** $$2 = 15(1 + 4k)$$ 8. **Expand the right side:** $$2 = 15 + 60k$$ 9. **Isolate $k$:** $$2 - 15 = 60k$$ $$-13 = 60k$$ 10. **Divide both sides by 60:** $$\cancel{60}k = \frac{-13}{\cancel{60}}$$ $$k = -\frac{13}{60}$$ **Final answer:** $$k = -\frac{13}{60}$$