Q:

Calculating $x$ from two equations involving $e^x$ and why is $\ln{(\frac{0.5}{0.1})}$ not the same as $\frac{\ln{0.5}}{\ln{0.1}}$

I have two equations as follows
$$0.1 = e^{-3μ}$$
$$0.5 = e^{-μx}$$
So I thought I would divide the second equation by the first
$$\frac{0.5}{0.1} = \frac{e^{-μx}}{e^{-3μ}}$$
I don't know how to go further. 
I thought I could take $\ln$ of both sides so 
$$\ln\left(\frac{0.5}{0.1}\right) = \frac{{-μx}}{-3μ}$$
but I don't think I'm right, or at least I have no idea why this is correct. 
Can someone explain if this is correct or not?
Update
Why is $$\ln\left(\frac{0.5}{0.1}\right)$$ not the same as $$\frac{\ln0.5}{\ln0.1}$$

A:

My methodology was wrong as 
$$\ln\left(\frac{0.5}{0.1}\right) \neq \left(\frac{\ln0.5}{\ln0.1}\right)$$
So to solve my equation instead, 
$$0.1 = e^{-3μ}$$
$$0.5 = e^{-μx}$$
I take logs of both sides of each equation
$$\ln0.1 = -3μ$$
$$\ln0.5 = -μx$$
Then divide the second equation by the first 
$$\frac{\ln0.5}{\ln0.1} = \frac{x}{3}$$
hence this gives me the final answer 
$$x = 3\left(\frac{\ln0.5}{\ln0.1}\right) = 0.903$$ to 3.s.f

