140Bquiz10csols - ii) False . The equation is (5 x ) = 11 /...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
1 MATH 140B QUIZ#10 Solutions Due: Nov, 2007 1. i) False . y = 2 ln( x 2 ) · 1 x 2 · (2 x ) = 4 ln( x 2 ) x = 8 ln x x . ii) False . y = 3 sin 2 p 1 2 ln 4 x P · cos p 1 2 ln 4 x P · 1 2 · 1 4 x · 4 , and so y = 3 sin 2 p 1 2 ln 4 x P · cos p 1 2 ln 4 x P · 1 2 x . iii) False y = e csc 2 (3 x ) · [ - 2 csc 2 (3 x ) cot(3 x )] · 3 Therefore, E) is the answer. 2. i) False . Since y = 1 2 ln(cos x ) - ln( e 2 x - 1) , we get y = - 1 2 tan x - 2 e 2 x e 2 x - 1 . ii) False . y = 1 e x - x · ( e x - 1) = e x - 1 e x - x . iii) True . Since y = 1 2 · ln x, we get y = - 1 2 x . Therefore, B) is the answer. 3. i) False . The domain is the set of all x such that x 2 - 1 > 0 . This is ( -∞ , - 1) (1 , ) . ii) False . Apply e to both sides giving x 2 + 2 = 3 x, which is the quadratic equation x 2 - 3 x + 2 = 0 , or ( x - 2)( x - 1) = 0. This gives x = 1 and x = 2 . iii) . You get the sequence of functions: 2 x log 2 x → - log 2 x → - log 2 ( x + 3) . Therefore, C) is the answer. 4. i) True . Take log 3 of both sides, move the 1 over, and divide by 2.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ii) False . The equation is (5 x ) = 11 / 4 . Taking log 5 of both sides gives x =-log 5 (11 / 4) = log 5 (4 / 11) . iii) False . This expression is not deFned. You cannot take the log of a negative number. Therefore, A) is the answer. 5. Taking ln of both sides and using the power rule gives, (3 x-1) ln e = (-5 x + 4) ln 2 . But ln e = 1 , so we have 3 x-1 =-(5 ln 2) x +4 ln 2 . Therefore (3+5 ln 2) x = 4 ln 2+1. Therefore, x = 4 ln 2 + 1 3 + 5 ln 2 . Therefore, E) is the answer....
View Full Document

This note was uploaded on 03/30/2008 for the course MATH 140B taught by Professor Fabbri,marc during the Fall '07 term at Pennsylvania State University, University Park.

Ask a homework question - tutors are online