We are asked to evaluate the triple integral $\int_{0}^{\log 2} \int_{0}^{x} \int_{0}^{x+y} e^{x+y+z} dz dy dx$.

1. Problem Description

We are asked to evaluate the triple integral $\int_{0}^{\log 2} \int_{0}^{x} \int_{0}^{x+y} e^{x+y+z} dz dy dx$.

2. Solution Steps

First, we evaluate the innermost integral with respect to $z$:

$\int_{0}^{x+y} e^{x+y+z} dz = e^{x+y} \int_{0}^{x+y} e^{z} dz = e^{x+y} [e^{z}]_{0}^{x+y} = e^{x+y} (e^{x+y} - e^{0}) = e^{x+y} (e^{x+y} - 1) = e^{2(x+y)} - e^{x+y}$.

Next, we evaluate the integral with respect to $y$:

$\int_{0}^{x} (e^{2(x+y)} - e^{x+y}) dy = \int_{0}^{x} e^{2x+2y} dy - \int_{0}^{x} e^{x+y} dy = e^{2x} \int_{0}^{x} e^{2y} dy - e^{x} \int_{0}^{x} e^{y} dy = e^{2x} [\frac{1}{2}e^{2y}]_{0}^{x} - e^{x} [e^{y}]_{0}^{x} = e^{2x} (\frac{1}{2}e^{2x} - \frac{1}{2}e^{0}) - e^{x} (e^{x} - e^{0}) = \frac{1}{2} e^{4x} - \frac{1}{2}e^{2x} - e^{2x} + e^{x} = \frac{1}{2} e^{4x} - \frac{3}{2}e^{2x} + e^{x}$.

Finally, we evaluate the integral with respect to $x$:

$\int_{0}^{\log 2} (\frac{1}{2} e^{4x} - \frac{3}{2}e^{2x} + e^{x}) dx = \frac{1}{2} \int_{0}^{\log 2} e^{4x} dx - \frac{3}{2} \int_{0}^{\log 2} e^{2x} dx + \int_{0}^{\log 2} e^{x} dx = \frac{1}{2} [\frac{1}{4}e^{4x}]_{0}^{\log 2} - \frac{3}{2} [\frac{1}{2}e^{2x}]_{0}^{\log 2} + [e^{x}]_{0}^{\log 2} = \frac{1}{8} (e^{4\log 2} - e^{0}) - \frac{3}{4} (e^{2\log 2} - e^{0}) + (e^{\log 2} - e^{0}) = \frac{1}{8} (e^{\log 2^4} - 1) - \frac{3}{4} (e^{\log 2^2} - 1) + (2 - 1) = \frac{1}{8} (16 - 1) - \frac{3}{4} (4 - 1) + 1 = \frac{15}{8} - \frac{9}{4} + 1 = \frac{15}{8} - \frac{18}{8} + \frac{8}{8} = \frac{15 - 18 + 8}{8} = \frac{5}{8}$.

3. Final Answer

5/8