Acnhduy Posted October 4, 2013 Share Posted October 4, 2013 (edited) Solve:log_{5}(x-1) + log_{5}(x-2) - log_{5}(x+6) = 0I feel like this is really simple but I cannot get the answer.This is what I attemptedSince they have the same base, I multiplied the toplog_{5}[(x-1)(x-2)] - log_{5}(x+6) = 0This gives melog_{5}(x^{2}-3x+2) - log_{5}(x+6) = 0At this point I did not now what to do, I tried 2 methods,1. Since they are the same base again, I divided the top this timelog_{5}(x^{2}-3x+2) / (x+6) = 0but I don't know how to continue2.I moved - log_{5}(x+6) to the other sidelog_{5}(x^{2}-3x+2) = log_{5}(x+6)Since they have the same base I ignored them sox^{2}-3x+2 = x+6Which gave mex^{2} - 4x - 4Which cannot be factored.Please help, thankyou Edited October 4, 2013 by Acnhduy Link to comment Share on other sites More sharing options...

studiot Posted October 4, 2013 Share Posted October 4, 2013 What methods, other than factorisation do you know for solving quadratic equations? Link to comment Share on other sites More sharing options...

Mozart Posted October 11, 2013 Share Posted October 11, 2013 There are basically three ways to solve a quadratic eguation. 1. by factorisation 2. completing the square method 3. using the almighty formula. I see no lapses with you manipulations. you are nearly there. Link to comment Share on other sites More sharing options...

studiot Posted October 11, 2013 Share Posted October 11, 2013 There are basically three ways to solve a quadratic eguation. FYI there is at least one more way. http://www.mathwarehouse.com/quadratic/roots/formula-sum-product-of-roots.php Link to comment Share on other sites More sharing options...

Mozart Posted October 14, 2013 Share Posted October 14, 2013 FYI there is at least one more way. http://www.mathwarehouse.com/quadratic/roots/formula-sum-product-of-roots.php FYI there is at least one more way. http://www.mathwarehouse.com/quadratic/roots/formula-sum-product-of-roots.php It's true. Thank you Studiot. I forgot that. Link to comment Share on other sites More sharing options...

ComplexCalcBro Posted October 14, 2013 Share Posted October 14, 2013 (edited) log_{5}(x-1) + log_{5}(x-2) - log_{5}(x+6) = 0 [(x-1)(x-2)]/(x+6) = 0 (x-1)(x-2) = 0 therefore: x = 1; 2 Since the denominator can never be zero. Edited October 14, 2013 by ComplexCalcBro Link to comment Share on other sites More sharing options...

overtone Posted October 14, 2013 Share Posted October 14, 2013 [(x-1)(x-2)]/(x+6) = 0 That is not correctly derived. Exp(0) = 1, not 0. The OP has it right, in the part where they "ignored the base". For many people, btw, the quadratic formula is more easily remembered by sound - memorized as a one would a short poem or limerick - rather than by sight. If you are one of those people, say it as you are writing it - by hand, with a pencil - a few times, and it will stick. Link to comment Share on other sites More sharing options...

Amaton Posted October 14, 2013 Share Posted October 14, 2013 (edited) [math]\log_5 (x-1)+\log_5 (x-2)-\log_5 (x+6)=0[/math] There are probably more efficient ways, but the first thing I thought of doing was... [math]5^{\log_5 (x-1)+\log_5 (x-2)-\log_5 (x+6)}=5^0[/math] Take 5 to the power of both sides. After some simplification... [math]\frac{(x-1)(x-2)}{x+6}=1[/math] From there, you can bring out the denominator to the other side and rearrange the equation into standard quadratic form. Edited October 14, 2013 by Amaton Link to comment Share on other sites More sharing options...

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