I would really appreciate a full solution with all working out to the following simultaneous equations, as I can't seem to arrive at the same answer as a text book on part of a question.
The equations are:
$$ V_1 = A \ln r_1 + B$$ $$ V_2 = A \ln r_2 + B$$
The answer in the book is:
$$ A = \frac{V_2 - V_1}{\ln r_2 - \ln r_1} = \frac{V_2 - V_1}{\ln (r_2/r_1)} $$
$$ B = \frac{V_1 \ln r_2 - V_2 \ln r_1}{\ln (r_2/r_1)} $$
Inserting these into the general equation $$ V(r) = A \ln r + B$$ gives:
$$ V(r) = \frac{V_2 - V_1}{\ln (r_2/r_1)} \ln r \ + \frac{V_1 \ln r_2 - V_2 \ln r_1}{\ln (r_2/r_1)} $$
$$ V(r) = \frac{V_2 \ln(r/r_1) - V_1 \ln(r/r_2)}{\ln (r_2/r_1)} $$
Thank you
Your equations are $$V_1=A\ln r_1+B\tag1$$ $$V_2=A\ln r_2+B\tag2$$ Subtracting $(1)$ from $(2)$ we obtain $$V_2-V_1=A(\ln r_2-\ln r_1)$$ $$\implies A=\frac{V_2 - V_1}{\ln r_2 - \ln r_1} = \frac{V_2 - V_1}{\ln (r_2/r_1)}$$ Pluging this result into $(1)$ we get $$\begin{align} V_1&=\frac{V_2 - V_1}{\ln (r_2/r_1)}\ln r_1 + B \\ \implies B&=V_1-\frac{V_2 - V_1}{\ln (r_2/r_1)}\ln r_1 \\ &= \frac{V_1 \ln(r_2/r_1)-(V_2-V_1)\ln r_1}{\ln (r_2/r_1)} \\ &=\frac{V_1\ln r_2-V_1\ln r_1-V_2\ln r_1+V_1\ln r_1}{\ln (r_2/r_1)} \\ &=\frac{V_1\ln r_2 - V_2\ln r_1}{\ln (r_2/r_1)} \\ \end{align}$$
Inserting these into the equation $V(r)=A\ln r+B$ gives $$\begin{align} V(r) &= \frac{V_2 - V_1}{\ln (r_2/r_1)} \ln r \ + \frac{V_1 \ln r_2 - V_2 \ln r_1}{\ln (r_2/r_1)} \\ &=\frac{V_2\ln r - V_1\ln r+V_1 \ln r_2 - V_2 \ln r_1}{\ln (r_2/r_1)} \\ &=\frac{V_2(\ln r - \ln r_1) - V_1(\ln r - \ln r_2)}{\ln (r_2/r_1)} \\ &=\frac{V_2\ln (r/r_1) - V_1\ln (r/r_2)}{\ln (r_2/r_1)} \end{align}$$ as desired.