This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: of equation you may have to solve. (b) Rewrite this to the form13 . 2 = log 10 x . Now take 1013 . 2 = 10 log 10 x = x . Therefore x = 6 . 3096 1014 . Note that the calculator notation ( 6.3095E14 ) is unacceptable . (c) Take a natural logarithm of both sides: ln 13 . 2 = ln e x 2 = x 2 . The answer falls out immediately: x = ln 13 . 2 = 1 . 6063. (d) Crossmultiply and rearrange: 4 . 3( x + 1) = x. 3 . 3 x =4 . 3 . x =1 . 3030 . 1 (e) Again, start by crossmultiplying and rearranging: a ( x + c ) = 2 x 2 + b. 2 x 2ax + bac = 0 . x = a p a 24(2)( bac ) 2(2) = 1 4 a p a 28( bac ) . 2...
View
Full
Document
This note was uploaded on 03/03/2012 for the course CHEM 2000 taught by Professor Roussel during the Fall '06 term at Lethbridge College.
 Fall '06
 Roussel
 pH

Click to edit the document details