Step 1: Identify the function and its components. The given function is $y = 8^{\frac{1}{3x}}$. This is an exponential function with base 8 and exponent $\frac{1}{3x}$. Step 2: Determine the domain of the exponent. The exponent $\frac{1}{3x}$ is defined for all real numbers $x$ except where the denominator is zero. $$ 3x \neq 0 \implies x \neq 0 $$ So, the domain of the function is $x \in (-\infty, 0) \cup (0, \infty)$. Step 3: Analyze the behavior of the exponent $k = \frac{1}{3x}$. • If $x > 0$, then $3x > 0$, so $k = \frac{1}{3x} > 0$. • If $x < 0$, then $3x < 0$, so $k = \frac{1}{3x} < 0$. The exponent $k$ can take any real value except 0. So, $k \in (-\infty, 0) \cup (0, \infty)$. Step 4: Determine the range of $y = 8^k$ based on the range of $k$. Since the base of the exponential function is $8$ (which is positive), $y = 8^k$ will always be positive. So $y > 0$. Consider two cases for $k$: Case 1: $k > 0$. If $k > 0$, then $8^k > 8^0$. Since $8^0 = 1$, this means $8^k > 1$. So, for $k \in (0, \infty)$, the range of $8^k$ is $(1, \infty)$. Case 2: $k < 0$. If $k < 0$, then $0 < 8^k < 8^0$. Since $8^0 = 1$, this means $0 < 8^k < 1$. So, for $k \in (-\infty, 0)$, the range of $8^k$ is $(0, 1)$. Step 5: Combine the ranges from both cases. The overall range of the function $y = 8^{\frac{1}{3x}}$ is the union of the ranges from Case 1 and Case 2. Range $= (0, 1) \cup (1, \infty)$. This means $y$ can be any positive real number except 1. The final answer is $\boxed{(0, 1) \cup (1, \infty)}$. 3 done, 2 left today. You're making progress.