Alg_Complete

# 6 6t ln 102 t ln16 0470003629246 3956 6 ln 102

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Unformatted text preview: ÷=3 è 1- x ø Now, convert it to exponential form. x2 - 6 x = 23 = 8 1- x Now, let’s solve this equation. x 2 - 6 x = 8 (1 - x ) x2 - 6 x = 8 - 8x x2 + 2 x - 8 = 0 ( x + 4 )( x - 2 ) = 0 Þ x = -4, x = 2 Now, let’s check both of these solutions in the original equation. x = -4 : © 2007 Paul Dawkins 306 http://tutorial.math.lamar.edu/terms.aspx College Algebra ( ) log 2 ( -4 ) - 6 ( -4 ) = 3 + log 2 (1 - ( -4 ) ) 2 log 2 (16 + 24 ) = 3 + log 2 ( 5 ) So, upon substituting this solution in we see that all the numbers in the logarithms are positive and so this IS a solution. Note again that it doesn’t matter that the solution is negative, it just can’t produce negative numbers or zeroes in the logarithms. x = 2: log 2 ( 22 - 6 ( 2 ) ) = 3 + log 2 (1 - 2 ) log 2 ( 4 - 12 ) = 3 + log 2 ( -1) In this case, despite the fact that the potential solution is positive we get negative numbers in the logarithms and so it can’t possibly be a solution. Therefore, we get a single solution fo...
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