-ln(0.1) equalling to ln(10)?

$\begingroup$

I am having quite a headache wrapping my head around this solution. I do not understand the first line where they get lambda = ln(10) from statement to the left. Somebody please explain this to me. Ignore the problem itself please.

$\endgroup$ 2

1 Answer

$\begingroup$

$$e^{- \lambda}=0.1 \Rightarrow e^{-\lambda}=\frac{1}{10} \Rightarrow \ln{e^{-\lambda}}=\ln{\frac{1}{10} } \Rightarrow -\lambda=\ln{1}-\ln{10} \Rightarrow -\lambda=0-\ln{10} \Rightarrow \\ \lambda=\ln{10}$$

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy