A compound interest continuously lets first set up

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Unformatted text preview: ege Algebra x ( x - 3) = 101 x 2 - 3x - 10 = 0 ( x - 5 )( x + 2 ) = 0 Þ x = -2, x = 5 So, we’ve got two potential solutions. Let’s check them both. x = -2 : log ( -2 ) = 1 - log ( -2 - 3) We’ve got negative numbers in the logarithms and so this can’t be a solution. x = 5: log 5 = 1 - log ( 5 - 3) log 5 = 1 - log 2 No negative numbers or zeroes in the logarithms and so this is a solution. Therefore, we have a single solution to this equation, x = 5 . Again, remember that we don’t exclude a potential solution because it’s negative or include a potential solution because it’s positive. We exclude a potential solution if it produces negative numbers or zeroes in the logarithms upon substituting it into the equation and we include a potential solution if it doesn’t. [Return to Problems] ( ) (c) log 2 x 2 - 6 x = 3 + log 2 (1 - x ) Again, let’s get the logarithms onto one side and combined into a single logarithm. log 2 ( x 2 - 6 x ) - log 2 (1 - x ) = 3 æ x2 - 6x ö log 2 ...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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