Solve for \( x \) \[ \log (x-1)-\log 2=\log x \]

To solve the equation \(\log (x-1) - \log 2 = \log x\), we can start by using the properties of logarithms. Recall that the difference of logs can be expressed like this: \[ \log (a) - \log (b) = \log \left(\frac{a}{b}\right) \] So, we can rewrite the left-hand side: \[ \log \left(\frac{x-1}{2}\right) = \log x \] Now, since the logarithms are equal, we can remove the logs (assuming \(x > 0\) and \(x - 1 > 0\)): \[ \frac{x-1}{2} = x \] Next, we can multiply both sides by 2 to eliminate the fraction: \[ x - 1 = 2x \] Now, we solve for \(x\) by getting all terms with \(x\) on one side: \[ -x = 1 \] \[ x = -1 \] However, we have to check if this value is valid in the original logarithmic equation. Since the logarithm is not defined for non-positive numbers, we find that \(x = -1\) is not valid. Now we need to check for other possible values. The initial substitution gives the condition \(x - 1 > 0 \Rightarrow x > 1\). Upon solving the modified equation again: From \(\log \left(\frac{x-1}{2}\right) = \log x\), we have \( \frac{x-1}{2} = x \) Solving yields: \[ x - 1 = 2x \Rightarrow -x = 1 \Rightarrow x = -1 \] This is indeed not valid due to the domain restrictions. Thus, upon conclusion, there are no real solutions for \(x\) that satisfy the original equation. So, the final answer is: There are no solutions for \( x \).