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Assignment 2 Solutions

# Assignment 2 Solutions - x(3 x-2 = 1 3 x 2-2 x-1 = 0 x-1(3...

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Assignment#2 Solutions Each question is worth 2, and the assignment is marked out of 10. 1. For what value of x does lb (2 x ) = log 4 ( x )? lb (2 x ) = log 4 ( x ) log 2 (2) + log 2 ( x ) = 1 2 log 2 ( x ) 1 2 log 2 ( x ) = - log 2 (2) 1 2 log 2 ( x ) = - 1 log 2 ( x ) = - 2 x = 2 - 2 x = 1 4 2. Solve for x : log 6 ( x + 4) + log 6 ( x - 3) = - 1. Firstly x > - 4 and x > 3 so x > 3. log 6 ( x + 4) + log 6 ( x - 3) = - 1 log 6 [( x + 4)( x - 3)] = - 1 ( x + 4)( x - 3) = 6 - 1 x 2 - x - 12 = 1 6 x 2 - x - 73 6 = 0 x = 1 ± q 1 2 - 4 * 1 * ( - 73 6 ) 2 x = 1 ± q 1 + ( 146 3 ) 2

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x = 1 ± q ( 149 3 ) 2 So x 4 . 02 or x ≈ - 3 . 02. By our ﬁrst condition, the second awnser is impossible. So the awnser is x 4 . 02. 3. How many digits in 3 100 ? First ﬁnd log 10 (3) = 0 . 4771212 ... . Then log 10 (3 100 ) = 100 log 10 (3) = 47 . 71212 ... Thus 47 log 10 (3 100 ) 48. Finally 10 47 3 100 10 48 . Since 10 47 has 48 digits, we know that 3 100 has 48 digits. 4. If f ( x ) = lb (4 x - 2), calculate in simplest form f ( x ) - f (1 - x ) f ( x ) - f (1 - x ) lb (4 x - 2) - lb (4(1 - x ) - 2) lb (4 x - 2) - lb (4 - 2 x ) lb (4 x - 2) (4 - 2 x ) lb 2(2 x - 1) 2(2 - x ) lb (2 x - 1) (2 - x ) 5. Solve for u : lg (lg ( u )) + lg (lg ( u 3 ) - 2) = 0. (Set x = lg ( u )) lg (lg ( u )) + lg (lg ( u 3 ) - 2) = 0 lg (lg ( u )) + lg (3 lg ( u ) - 2) = 0 lg ( x ) + lg (3 x - 2) = 0 lg [( x )(3 x - 2)] = 0

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Unformatted text preview: ( x )(3 x-2) = 1 3 x 2-2 x-1 = 0 ( x-1)(3 x + 1) = 0 So x = 1 ,-1 3 . lg ( u ) = 1 u = 10 1 = 10 lg ( u ) =-1 3 u = 10-1 3 = 1 3 √ 10 So u = 10 , 1 3 √ 10 . 6. Find x : x 1-1 2 (lg x ) 2-1 2 √ 10 = 0 First of all, x > 0. x 1-1 2 (lg x ) 2-1 2 √ 10 = 0 x 1-1 2 (lg x ) 2 = 1 2 √ 10 log 10 [ x 1-1 2 (lg x ) 2 ] = log 10 1 2 √ 10 [1-1 2 (log 10 x ) 2 ] log 10 ( x ) =-1 2 Set y = log 10 ( x ). y-1 2 y 3 =-1 2 y 3-2 y-1 = 0 By remainder theorem, -1 is a root. y 3-2 y-1 = ( y + 1)( y 2-y-1) By quadratic formula, the other two roots are: y = 1 ± p (-1) 2-4 * 1 * (-1) 2 y = 1 ± √ 5 2 Therefore: log 10 ( x ) =-1 x = 1 10 OR log 10 ( x ) = 1 ± √ 5 2 x = 10 1 ± √ 5 2 x = √ 10 * 10 ± √ 5 2 So we get three possible x values....
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Assignment 2 Solutions - x(3 x-2 = 1 3 x 2-2 x-1 = 0 x-1(3...

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