#(dL)/L = \alphadT + 1/(sE_T)df \rightarrow ln(L/L_0) = \alpha\DeltaT + f/(sE_T)# Is that mathematically correct ?

1 Answer
Cesareo R.
Oct 24, 2017

See below.

Explanation:

Considering an expression as

#L = e^(alphaT+f/(sE_T))# applying #log# to both sides

#logL = alphaT+f/(sE_T)# now deriving totally

#(dL)/L = alpha dT +(df)/(sE_T)# integrating we have

#int_(L_0)^L(dL)/L = int_(T_0)^T alpha dT +int_(f_0)^f 1/(sE_T) df# or

#log(L/L_0) = alpha (T-T_0) + (f-f_0)/(sE_T)#

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