Spoiler: Is this correct? Cause the texbook gives. $\displaystyle (b^{b}a^{-a})^{\frac{1}{b-a}}e^{-1}$ The original question is at the bottom on the picture i posted.
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Yes thats correct. You can simplify your answer to the one given in the textbook. Remember that $\displaystyle e^{\ln x} = x$ $\displaystyle \, \, $ !
correction x not x!, sry, just had to do it Originally Posted by Isomorphism Yes thats correct. You can simplify your answer to the one given in the textbook. Remember that $\displaystyle e^{\ln x} = x$!
Last edited by matheagle; May 20th 2009 at 10:36 PM.
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