how to solve for q in the equation y=log(-log(1-q))

y = log(-log(1-q))

Raise the terms with the base 10.

10^y = 10^{log(-log(1-q))}

Now, this cancels the first log.

10^y = -log(1-q)

Move the minus sign, to become the power of (1-q)

10^y = log(1-q)^{-1}

Take the base 10 again:

10^{10^y} = 10^{log(1-q)^{-1}}

This takes out the other log.

10^{10^y} = (1-q)^{-1}

Convert the right side to a fraction:

10^{10^y} = \frac{1}{1-q}

Can you finish it now? Post what you get! :)

Please do not double post the same question in two different forums. See: https://www.askmehelpdesk.com/mathematics/how-solve-q-equation-y-log-log-1-q-513930.html

It is interesting though that Galactus and Unknown008 got the answer in two different forms - though they are eqiuvalent.

Yup, I like removing negative signs as soon as possible ;)

I wonder how I missed the other thread :rolleyes:

Must be tired >_<