How does "this" equation turn into "that" equation?

[calmheart]

Hello friends, I am having trouble understanding how equation 1 becomes equation 2…

\(\displaystyle \mbox{Eqn 1. }\, K\, \times\, \left(e^{g\tau}\, \times\, e^{-g\tau}\right)\, \times\, e^{gT}\, \times\, e^{-rT}\, \times\, e^{r\tau}\)

\(\displaystyle \mbox{Eqn 2. }\, K\, \times\, e^{g\tau}\,\left(e^{-rT}\, \times\, e^{gT}\, \times\, e^{r\tau}\, \times\, e^{-g\tau}\right)\)

 

[stapel]

Hello friends, I am having trouble understanding how equation 1 becomes equation 2…

\(\displaystyle \mbox{Eqn 1. }\, K\, \times\, \left(e^{g\tau}\, \times\, e^{-g\tau}\right)\, \times\, e^{gT}\, \times\, e^{-rT}\, \times\, e^{r\tau}\)

\(\displaystyle \mbox{Eqn 2. }\, K\, \times\, e^{g\tau}\,\left(e^{-rT}\, \times\, e^{gT}\, \times\, e^{r\tau}\, \times\, e^{-g\tau}\right)\)

Had there been another "times" symbol in front of the parenthetical in Eqn. 2, these would have been equivalent, having only had their terms rearranged. (Remember that multiplication is commutative.) However, lacking that symbol, the term immediately in front of the parenthetical is carried through onto all the terms inside, yielding:

. . . . .\(\displaystyle \mbox{Eqn 2. }\, K\, \times\, e^{g\tau-rT}\,\times\, e^{g\tau+gT}\, \times\, e^{g\tau+r\tau}\, \times\, e^{g\tau-g\tau}\)

The last term above, of course, simplifies to just "1". So, unless you've been given a bunch of other information regarding the variables, etc, the two lines are not equivalent.

If you have been told that they are, you might want to suggest that there is a "times" symbol missing. (That is, that the text contains a typo.)