• Support PF! Buy your school textbooks, materials and every day products Here!

Expanding ln isn't working for me

  • Thread starter iScience
  • Start date
  • #1
465
4
original equation: ln(0.2048x[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?
 

Answers and Replies

  • #2
Pythagorean
Gold Member
4,191
255
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.
 
  • #3
465
4
oops, lol thanks
 
  • #4
Pythagorean
Gold Member
4,191
255
no problem
 

Related Threads on Expanding ln isn't working for me

  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
3
Views
60K
  • Last Post
Replies
3
Views
18K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
23
Views
1K
  • Last Post
Replies
3
Views
9K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
4
Views
7K
Top