1. **State the problem:** Given $y = \log(2t)$ and $x = e^{-2t}$, prove that $$\frac{d^2y}{dx^2}\bigg|_{t=1} = \frac{e^4}{4}.$$\n\n2. **Recall the chain rule for derivatives:** To find $\frac{d^2y}{dx^2}$, we use \n$$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}.$$\n\n3. **Find $\frac{dy}{dt}$ and $\frac{dx}{dt}$:**\n- $y = \log(2t)$ so $$\frac{dy}{dt} = \frac{1}{2t} \cdot 2 = \frac{1}{t}.$$\n- $x = e^{-2t}$ so $$\frac{dx}{dt} = -2e^{-2t}.$$\n\n4. **Find $\frac{dy}{dx}$ using the chain rule:**\n$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{1}{t}}{-2e^{-2t}} = -\frac{e^{2t}}{2t}.$$\n\n5. **Differentiate $\frac{dy}{dx}$ with respect to $t$ to find $\frac{d}{dt}\left(\frac{dy}{dx}\right)$:**\n$$\frac{d}{dt}\left(-\frac{e^{2t}}{2t}\right) = -\frac{d}{dt}\left(\frac{e^{2t}}{2t}\right).$$\nUse the quotient rule or product rule: write as $$-\frac{1}{2} \cdot \frac{e^{2t}}{t}.$$\n\nLet $$u = e^{2t}, \quad v = t,$$ then\n$$\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} = \frac{2e^{2t} \cdot t - e^{2t} \cdot 1}{t^2} = \frac{e^{2t}(2t - 1)}{t^2}.$$\n\nSo,\n$$\frac{d}{dt}\left(-\frac{e^{2t}}{2t}\right) = -\frac{1}{2} \cdot \frac{e^{2t}(2t - 1)}{t^2} = -\frac{e^{2t}(2t - 1)}{2t^2}.$$\n\n6. **Find $\frac{dt}{dx}$:**\nSince $$\frac{dx}{dt} = -2e^{-2t},$$ then\n$$\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = -\frac{1}{2} e^{2t}.$$\n\n7. **Calculate $\frac{d^2y}{dx^2}$:**\n$$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx} = \left(-\frac{e^{2t}(2t - 1)}{2t^2}\right) \cdot \left(-\frac{1}{2} e^{2t}\right) = \frac{e^{4t}(2t - 1)}{4t^2}.$$\n\n8. **Evaluate at $t=1$:**\n$$\frac{d^2y}{dx^2}\bigg|_{t=1} = \frac{e^{4 \cdot 1}(2 \cdot 1 - 1)}{4 \cdot 1^2} = \frac{e^4 (2 - 1)}{4} = \frac{e^4}{4}.$$\n\n**Final answer:** $$\boxed{\frac{d^2y}{dx^2}\bigg|_{t=1} = \frac{e^4}{4}}.$$