Solving for $x$ using $\ln$ I have an equation as follows:

$\ln(a)x + \ln(b) (1 - x) = \ln(c)$

I'd like to solve for $x$, but I think what I've come up with is wrong

$\ln\left(\frac{a}{b}\right)x + \ln(b) = \ln(c)$

$\ln\left(\frac{a}{b}\right)x = ln\left(\frac{c}{b}\right)$

$x = \frac{\ln\left(\frac{c}{b}\right)}{\ln\left(\frac{a}{b}\right)}$
$$x\ln a+(1-x)\ln b=\ln c$$
$$x\log_ba+1-x=\log_bc$$ 
$$x(\log_ba-1)=\log_bc-1$$
$$x=\frac{\log_bc-1}{\log_ba-1}=\frac{\log_b(\frac cb)}{\log_b(\frac ab)}$$
$$x\ln a+(1-x)\ln b=\ln c$$
$$x\log_ba+1-x=\log_bc$$ 
$$x(\log_ba-1)=\log_bc-1$$
$$x=\frac{\log_bc-1}{\log_ba-1}=\frac{\log_b(\frac cb)}{\log_b(\frac ab)}$$