Solving $y_1=a\ln(b x_1)$ and $y_2=a\ln(b x_2)$ for $a$ and $b$ Maths found from another post:

How is a and b equated? I don't understand the method and am hoping someone is able to help and show the process.
 A: We want to solve the system
$$
\left\{
\begin{array}{lcl}
y_1 & = & a \ln(bx_1) \\
y_2 & = & a \ln(bx_2) 
\end{array}
\right.
$$
The key point is to notice that, once $a$ is determined, we can easily obtain $b$ since
$$
y=a\ln(ax) \iff \frac{y}{a}=\ln(bx) \iff bx = \exp\left(\frac{y}{a}\right) \iff b = \frac{\exp(y/a)}{x}
$$
To find $a$, assume first that $x_1\neq x_2$. Subtracting the two equations, we get
$$y_1-y_2= a\left(\ln(bx_1)-\ln(bx_2)\right)=a\ln\left(\frac{bx_1}{bx_2}\right)=a\ln\left(\frac{x_1}{x_2}\right)$$
so $$a=\frac{y_1-y_2}{\ln\left(\frac{x_1}{x_2}\right)}
=\frac{y_1-y_2}{\ln(x_1)-\ln(x_2)}$$
This implies that
$$
\frac{y_1}{a} = \frac{y_1\ln(x_1)-y_1\ln(x_1)}{y_1-y_2}
$$
so
$$
b = \frac{\exp\left(\frac{y_1\ln(x_1)-y_1\ln(x_1)}{y_1-y_2}\right)}{x_1}
$$
To obtain the formula from the answer in the OP, we can write $x_1=\exp(\ln(x_1))$ and we get
$$
b 
= 
\frac{\exp\left(\frac{y_1\ln(x_1)-y_1\ln(x_1)}{y_1-y_2}\right)}{\exp(\ln(x_1))}
=
\exp\left(\frac{y_1\ln(x_1)-y_1\ln(x_1)}{y_1-y_2}-\ln(x_1)\right)
$$
Finally,
$$
b=
\exp\left(\frac{y_1\ln(x_1)-y_1\ln(x_1)-(y_1-y_2)\ln(x_1)}{y_1-y_2}\right)
=
\exp\left(\frac{y_1\ln(x_2)-y_2\ln(x_1)}{y_1-y_2}\right)
$$
If $x_1=x_2$, then $y_1\neq y_2$, so the system has no solution.
