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Unformatted text preview: Version 57 1. Solve the inequality ln( x 2) + ln( x 5) ln( x 4) < ln(3) . Give your answer in interval notation. Make sure that the equation actually makes sense for the interval you give in your answer. Solution: We exponentiate to get ( x 2)( x 5) ( x 4) < 3 . We can assume that x 4 > for otherwise ln( x 4) will not make sense. It is therefore OK to multiply by x 4 . This gives ( x 2)( x 5) < 3( x 4) or x 2 7 x + 10 < 3 x 12 or x 2 10 x + 22 < or ( x 5) 2 25 + 22 < or  x 5  2 < 3 or 5 3 < x < 5 + 3 . But the original equation only makes sense for x > 5 and 5 3 < 5 , so the actual range is 5 < x < 5 + 3 . In interval notation (5 , 5 + 3) . 2. Find tan( x ) where x is the smallest solution of 3 tan( x ) + 7 tan x + 4 = 5 in the range 2 < x < 2 . Solution: We have tan x + 4 = tan( x ) + tan ( 4 ) 1 tan( x ) tan ( 4 ) = tan( x ) + 1 1 tan( x ) since tan 4 = 1 . Therefore, setting t = tan( x ) we have 3 t + 7 t + 1 1 t =...
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This note was uploaded on 01/11/2011 for the course MATH MATH 140 taught by Professor Drury during the Fall '10 term at McGill.
 Fall '10
 Drury

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