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

Question

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 $$\frac{\ln0.5}{\ln0.1}$$

Answer 1

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

Answer 2

Your simplification is wrong : $$\ln(\frac{a}{b}) \neq \frac{\ln{a}}{\ln{b}}$$ However your intuition was good : $$\frac{0.5}{0.1}=\frac{e^{-\mu x}}{e^{-3\mu}}$$ $$5=e^{-\mu(x-3)}$$ $$x=3-\frac{\ln(5)}{\mu}$$