Bryn Posted June 10, 2004 Share Posted June 10, 2004 how would you go about solving [math]log_3x - 2log_x3 = 1[/math] Link to comment Share on other sites More sharing options...
e(ho0n3 Posted July 19, 2004 Share Posted July 19, 2004 Make the substitution [math]\log_x{3} = \log_3{3}/\log_3{x}[/math] Manipulate the equation until you get [math](\log_3{x})^2 - \log_3{x} - 2 = 0[/math] Then make the substitution [math]u = \log_3{x}[/math] Solve for u, then solve for x. Link to comment Share on other sites More sharing options...
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