If Laplace transform \(Lf\left( t \right) = \log \left( {\frac{{s + a}}{{s + b}}} \right)\), then f(t) equals

\(\frac{1}{t}\left( {{e^{bt}} - {e^{at}}} \right)\)

\(\frac{1}{t}\left( {{e^{ - bt}} - {e^{ - at}}} \right)\)

If Laplace transform \(Lf\left( t \right) = \log \left( {\frac{{s + a

1.	If Laplace transform \(Lf\left( t \right) = \log \left( {\frac{{s + a}}{{s + b}}} \right)\), then f(t) equals
A.	\(\frac{1}{t}\left( {{e^{ - bt}} - {e^{ - at}}} \right)\)
B.	\(\frac{1}{t}\left( {{e^{bt}} - {e^{at}}} \right)\)
C.	\({e^{ - bt}} - {e^{ - at}}\)
D.	\({e^{bt}} - {e^{at}}\)
Answer» B. \(\frac{1}{t}\left( {{e^{bt}} - {e^{at}}} \right)\)