1. **State the problem:** We are given the equation $\log JC - \log (2x-1) = 1$ and need to find the value of $x$. 2. **Recall the logarithm subtraction rule:** The difference of logarithms with the same base can be written as the logarithm of a quotient: $$\log a - \log b = \log \left(\frac{a}{b}\right)$$ 3. **Apply the rule to the equation:** $$\log JC - \log (2x-1) = \log \left(\frac{JC}{2x-1}\right) = 1$$ 4. **Rewrite the logarithmic equation in exponential form:** Assuming the logarithm base is 10 (common log), $$\log \left(\frac{JC}{2x-1}\right) = 1 \implies \frac{JC}{2x-1} = 10^1 = 10$$ 5. **Solve for $x$:** $$\frac{JC}{2x-1} = 10 \implies JC = 10(2x-1)$$ $$JC = 20x - 10$$ $$20x = JC + 10$$ $$x = \frac{JC + 10}{20}$$ **Final answer:** $$x = \frac{JC + 10}{20}$$