Question

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

Answer 1

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)`