# Question #47835

Mar 15, 2017

#### Explanation:

First we notice that this is a quotient so logically we must use the quotient rule to solve this. The quotient rule generically looks like

$f \frac{x}{g} \left(x\right) = \left(\frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g {\left(x\right)}^{2}}\right)$

And in this case we would use $\ln \left(6 x\right) = f \left(x\right) \mathmr{and} \ln \left(2 x\right) = g \left(x\right)$

So to solve this equation we would want to find the derivative of both the top and the bottom

$f ' \left(x\right) = \frac{6}{6 x} = \frac{1}{x} \mathmr{and} g ' \left(x\right) = \frac{2}{2 x} = \frac{1}{x}$ and plug everything into the quotient rule would look like the following

$\frac{\left(\ln \left(2 x\right) \left(\frac{1}{x}\right)\right) - \left(\ln \left(6 x\right) \left(\frac{1}{x}\right)\right)}{\ln {\left(2 x\right)}^{2}}$

Which we can simplify to

$\left(\frac{\frac{\ln \left(2 x\right) - \ln \left(6 x\right)}{x}}{\ln {\left(2 x\right)}^{2}}\right)$

Simplified even further looks like

$\left(\frac{\ln \left(2 x\right) - \ln \left(6 x\right)}{x \left(\ln {\left(2 x\right)}^{2}\right)}\right)$