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Review3 Solutions - Review Sheet 3 Solutions Math 112 Fall...

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Review Sheet 3 Solutions Math 112 Fall 2008 1. Graph y = 2 x +1 - 3 This graph will have an asymptote of x = - 3 and intercepts (0 , - 1) and (log 2 3 - 1 , 0). 2. Graph y = - log( x - 2) + 2 This graph will have an asymptote of x = 2 and intercepts (0 , - log( - 2) + 3) and (102 , 0). 3. Graph y = e - x +2 - 1 This graph will have asymptote of y = - 1 and intercepts (0 , e 2 - 1) and (2 , 0). 4. Write the expressions as a single logarithm. A) log(6) + log(1 / 3) + log(10) log(20) B) ln(2) - 4(ln(5) - ln(4)) ln 512 625 C) log(4) + 3[log( x - 1) - 2 log( x 2 - 1)] log 4( x - 1) 3 ( x 2 - 1) 6 = log 4 ( x - 1) 3 ( x + 1) 6 5. Write the quantity using sums and differences of the simplest possible logs. A) log x 2 - 4 1 2 log( x + 4) + 1 2 log( x - 4) 1
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2 B) log 3 q x - 1 2 x +3 1 3 log( x - 1) - 1 3 log(2 x + 3) 6. Write the quantity in terms of ln( x ) and ln( y ): ln xy + ln( x/e 2 ) - ln( e 2 x y ) 1 2 ln( x ) + 1 2 ln( y ) + ln( x ) - 2 - 2 - ln( x ) - 1 2 ln( y ) = 1 2 ln( x ) - 4 7. Solve the following equations: A) log(2 x 2 - 4) = 5 2 x 2 - 4 = 100000 2 x 2 = 100004 x 2 = 50002 x = ± 50002 B) (ln( x )) 3 = 3 ln( x ) Letting t = ln( x ), we get that t ( t 2 - 3) = 0 and hence t = 0 , ± 3 We disregard the first solution, hence we get x = e 3 , x = e - 3 C) ln(4) - ln( x ) = (ln(4)) / (ln( x )) Again, let t = ln( x ). Then we have t ln(4) - t 2 = ln(4) and hence t = - ln(4) ± p (ln(4)) 2 + 4 ln(4) 2 But this has complex solutions, which we disregard. Hence there is no solution. D) log( x + 1) = 2 + log( x - 1)
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3 log( x + 1) - log( x - 1) = 2 log x + 1 x - 1 = 2 x + 1 x - 1 = 100 x + 1 - 100 x + 100 x - 1 = 0 - 99 x + 101 x - 1 = 0 - 99 x + 101 = 0 99 x = 101 x = 101 99 This answer will result in a positive number inside both of the original logarithms, so it is a solution.
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