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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online