Solving $y=a\log\left(bx\right)$ using the least squares method

Question

How do I solve for $a$ and $b$ in $y=a\log\left(bx\right)$ using the method of least squares? I tried using $$10^y = \left(bx\right)^a$$ but I end up with terms containing both $a$ and $b$.

Answer 1

$$ y = a\log{bx} = a\log{b} + a\log{x} = \left| \begin{aligned} \beta_0 &= a\log{b} \\ \beta_1 &= a \\ X &= \log{x} \end{aligned} \right| = \beta_0 + \beta_1 X $$

You can define estimates $\beta_0, \beta_1$ by the least squares method.

Then,

$$ \begin{aligned} a &= \beta_1 \\ \beta_0 &= a\log{b} \Leftrightarrow \beta_0 = \beta_1\log{b} \Leftrightarrow \log{b} = \frac{\beta_0}{\beta_1} \Leftrightarrow b = e^\frac{\beta_0}{\beta_1} \end{aligned} $$

Answer 2

Hint

$$y=a\log\left(bx\right)\implies y=a(log(b)+\log(x))=a\log(b)+a\log(x)$$ Define $c=a\log(b)$ and $z=\log(x)$.

That's it