I already know the value of y, can you help me to find T? thanks :)

Question

$y = - ln (- ln (\frac{T - 1}{T}))$ $y=-ln(-ln((T-1)/T))$

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Answer 1 · 2017-11-23T04:41:26

$T = \frac{1}{1 - e^{- e^{- y}}}$ $T=1/(1-e^(-e^-y))$

Okay, let's solve this step by step.

$y = - ln (- ln (\frac{T - 1}{T}))$ $y=−ln(−ln((T−1)/T))$

$- y = ln (- ln (\frac{T - 1}{T}))$ $-y=ln(-ln((T-1)/T))$

To make things simpler, let's set $a = - ln (\frac{T - 1}{T})$ $a=-ln((T-1)/T)$

$- y = ln (a)$ $-y=ln(a)$

This equation basically means:

$e^{- y} = a$ $e^-y=a$

Let's continue:

$e^{- y} = - ln (\frac{T - 1}{T})$ $e^-y=-ln((T-1)/T)$

$- e^{- y} = ln (\frac{T - 1}{T})$ $-e^-y=ln((T-1)/T)$

Once again, we can set a variable for $\frac{T - 1}{T}$ $(T-1)/T$ and get rid of the logarithm. Eventually, we will get:

$e^{- e^{- y}} = \frac{T - 1}{T}$ $e^(-e^-y)=(T-1)/T$

$T \cdot e^{- e^{- y}} = T - 1$ $T*e^(-e^-y)=T-1$

$1 = T - T \cdot e^{- e^{- y}}$ $1=T-T*e^(-e^-y)$

$T (1 - e^{- e^{- y}}) = 1$ $T(1-e^(-e^-y))=1$

$T = \frac{1}{1 - e^{- e^{- y}}}$ $T=1/(1-e^(-e^-y))$

Since you know $y$ $y$ , you can solve for $T$ $T$ . If you see any mistakes, just comment. Hope this helps!