# Expanding ln isn't working for me

original equation: ln(0.2048x$\frac{5}{2}$)= -$\frac{5}{2}$

when i just raise this to e i get..

0.2048T$\frac{5}{2}$= -$\frac{5}{2}$

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

ln(0.2048) + $\frac{5}{2}$ln(x)= -$\frac{5}{2}$

0.2048+$\frac{5}{2}$x=e-$\frac{5}{2}$

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

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Pythagorean
Gold Member
you can't do that with a sum of ln's. Taking the exponent of both sides after expansion does not cancel out the ln in each term. You would have e^(lnA + lnB) and that does not equal A+B.

oops, lol thanks

Pythagorean
Gold Member
no problem