What is ln(1/((3.5e^{(-4T))-5e^{(-T)}+2.5)}) ?

The solution to ln(1/((3.5e^{(-4T))-5e^{(-T)}+2.5)}) is -ln(3.5e^{-4T}-5e^{-T}+2.5)

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simplify ln(1/((3.5e^{(-4T))-5e^{(-T)}+2.5)})

simplify

ln (\frac{1}{( 3.5 e ^{(- 4 T)} - 5 e ^{(- T)} + 2.5 )})

- ln (3.5 e^{- 4 T} - 5 e^{- T} + 2.5)

Solution steps

ln (\frac{1}{( 3.5 e ^{(- 4 T)} - 5 e ^{(- T)} + 2.5 )})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{1}{( 3.5 e ^{(- 4 T)} - 5 e ^{(- T)} + 2.5 )}) = ln (1) - ln ((3.5 e^{(- 4 T)} - 5 e^{(- T)} + 2.5))

= ln (1) - ln (3.5 e^{- 4 T} - 5 e^{- T} + 2.5)

Apply log rule:

lo g_{a} (1) = 0

= 0 - ln (3.5 e^{- 4 T} - 5 e^{- T} + 2.5)

0 - ln (3.5 e^{- 4 T} - 5 e^{- T} + 2.5) = - ln (3.5 e^{- 4 T} - 5 e^{- T} + 2.5)

= - ln (3.5 e^{- 4 T} - 5 e^{- T} + 2.5)

lo g_{b} (\frac{a ^{2} 2}{e ^{3}})

ln (v^{- \frac{1}{3}}) + ln ((v - 3)^{\frac{1}{3}}) + c

13 lo g_{10} (a) - lo g_{10} (b)

ln (\frac{x ^{2}}{x ^{2} + 1}) - \frac{1}{x} - arctan (x) + c

1 - ln (\frac{3}{2})

What is ln(1/((3.5e^{(-4T))-5e^{(-T)}+2.5)}) ?
The solution to ln(1/((3.5e^{(-4T))-5e^{(-T)}+2.5)}) is -ln(3.5e^{-4T}-5e^{-T}+2.5)