\begin{align} \chi^2 &= \sum_i \left[ \ln(y_i-c) - \ln a +bx_i \right]^2\\ &= \sum_i \left[ \left[ \ln(y_i-c) \right]^2 + \left(\ln a\right)^2 + b^2 x_i^2 - 2 \ln a\ln(y_i-c) + 2 bx_i\ln(y_i-c) - 2b x_i \ln a\right]\\ \end{align}

\begin{align} \frac{\partial\chi^2}{\partial \ln a} &= \sum_i \left[ 2 \ln a - 2 \ln(y_i-c) - 2b x_i \right] = 0\\ \frac{\partial\chi^2}{\partial b} &= \sum_i \left[ 2b x_i^2 + 2 x_i \ln\left( y_i-c\right) -2 x_i \ln a \right]= 0\\ \frac{\partial\chi^2}{\partial c} &= \sum_i \left[ 2 \frac{\ln(y_i-c)}{y_i-c} -2 \frac{\ln a}{y_i-c} + 2\frac{b x_i}{y_i-c}\right]= 0\\ \end{align}

\begin{align} \sum_i \left[ \ln a - \ln(y_i-c) - b x_i \right] = 0\\ \sum_i \left[ b x_i^2 + x_i \ln\left( y_i-c\right) - x_i \ln a \right] = 0\\ \sum_i \left[ \frac{\ln(y_i-c)}{y_i-c} - \frac{\ln a}{y_i-c} + \frac{b x_i}{y_i-c}\right] = 0\\ \end{align}

\begin{align} n\ln a - \sum_i \ln(y_i-c) - b \sum_i x_i = 0\\ b \sum_i x_i^2 + \sum_i x_i \ln\left( y_i-c\right) - \ln a \sum_i x_i = 0\\ \sum_i \frac{\ln(y_i-c)}{y_i-c} - \ln a \sum_i \frac{1}{y_i-c} + b \sum_i \frac{x_i}{y_i-c}= 0\\ \end{align}