1. **State the problem:** We are given the exponential function $$y = 300 \left(\frac{1}{2}\right)^{-\frac{x}{68}}$$ and need to determine if it represents exponential growth or decay, and find the percentage rate of increase or decrease per unit of $x$ to the nearest tenth of a percent. 2. **Identify the form and base:** The function is of the form $$y = a b^x$$ where $$a = 300$$ and $$b = \left(\frac{1}{2}\right)^{-\frac{1}{68}}$$ because the exponent is $$-\frac{x}{68} = x \cdot \left(-\frac{1}{68}\right)$$. 3. **Simplify the base:** Using the property of exponents $$b = \left(\frac{1}{2}\right)^{-\frac{1}{68}} = 2^{\frac{1}{68}}$$ because $$\left(\frac{1}{2}\right)^{-1} = 2$$. 4. **Determine growth or decay:** Since $$b = 2^{\frac{1}{68}} > 1$$, the function represents exponential growth. 5. **Calculate the growth rate:** The growth rate per unit $x$ is $$r = b - 1 = 2^{\frac{1}{68}} - 1$$. 6. **Evaluate the growth rate numerically:** Calculate $$2^{\frac{1}{68}} = e^{\frac{1}{68} \ln 2} \approx e^{0.0102} \approx 1.0103$$. So, $$r \approx 1.0103 - 1 = 0.0103$$ or 1.03% growth per unit $x$. 7. **Round to nearest tenth of a percent:** The growth rate is approximately **1.0%** per unit $x$. **Final answer:** The function represents exponential growth at a rate of approximately 1.0% per unit increase in $x$.