- #1

iScience

- 466

- 5

^{[itex]\frac{5}{2}[/itex]})= -[itex]\frac{5}{2}[/itex]

when i just raise this to e i get..

0.2048T

^{[itex]\frac{5}{2}[/itex]}= -[itex]\frac{5}{2}[/itex]

and then i can solve for x and get the right answer. but when i expand the logs..

ln(0.2048) + [itex]\frac{5}{2}[/itex]ln(x)= -[itex]\frac{5}{2}[/itex]

0.2048+[itex]\frac{5}{2}[/itex]x=e

^{-[itex]\frac{5}{2}[/itex]}

then i try to solve for x i don't get the right answer. what am i doing wrong?