# HW7 - Math 191 HW 7 Solutions 7.4 6 a log 9 x log 3 x = ln...

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Unformatted text preview: Math 191 HW 7 Solutions 7.4: 6 a. log 9 x log 3 x = ln x ln 9 Ã· ln x ln 3 = ln x 2 ln 3 Â· ln 3 ln x = 1 2 b. log âˆš 10 x log âˆš 2 x = ln x ln âˆš 10 Ã· ln x ln âˆš 2 = ln x 1 2 ln 10 Â· 1 2 ln 2 ln x = ln 2 ln 10 c. log a b log b a = ln b ln a Ã· ln a ln b = ln b ln a Â· ln b ln a = ( ln b ln a ) 2 7.4: 8 8 log 8 (3)- e ln 5 = x 2- 7 log 7 (3 x ) â‡’ 3- 5 = x 2- 3 x â‡’ 0 = x 2- 3 x + 2 = ( x- 1)( x- 2) â‡’ x = 1 or x = 2 7.4: 20 y = 3 tan Î¸ ln 3 â‡’ dy dÎ¸ = (3 tan Î¸ ln 3)(ln3) sec 2 Î¸ = 3 tan Î¸ (ln 3) 2 sec 2 Î¸ 7.4: 64 R e 1 2 ln 10(log 1 x ) x dx = R e 1 (ln 10)(2 ln x ) (ln 10) 1 x dx = [(ln x ) 2 ] e 1 = (ln e ) 2- (ln 1) 2 = 1 7.4: 72 R e x 1 1 t dt = [ln | t | ] e x 1 = ln e x- ln 1 = x ln e = x 7.5: 6 V ( t ) = V e- t/ 40 â‡’ . 1 V = V e- t/ 40 when the voltage is 10% of its original value â‡’ t =- 40 ln(0 . 1) â‰ˆ 92 . 1 sec 7.5:8 y = y e kt and y (3) = 10 , 000 â‡’ 10 , 000 = y e 3 k ; also y (5) = 40 , 000 = y e 5 k . Therefore y e 5 k = 4 y e 3 k â‡’ e 5 k = 4 e 3 k...
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## This note was uploaded on 06/19/2008 for the course MATH 1910 taught by Professor Berman during the Spring '07 term at Cornell.

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HW7 - Math 191 HW 7 Solutions 7.4 6 a log 9 x log 3 x = ln...

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