# SOLUTION: How do I solve the following problem: logx-log3=log4-log(x+4).

Algebra ->  Algebra  -> Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How do I solve the following problem: logx-log3=log4-log(x+4).       Log On

 Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Algebra: Logarithm Solvers Lessons Answers archive Quiz In Depth

 Question 28162: How do I solve the following problem: logx-log3=log4-log(x+4). Answer by sdmmadam@yahoo.com(530)   (Show Source): You can put this solution on YOUR website!logx-log3=log4-log(x+4). ----(1) logx+log(x+4)= log4 +log3 [adding log(x+4)and log3 to both the side) log[x(x+4)]= log(4X3) [using log(a) + log(b) = log(ab)(of course when the base is the same)] [x(x+4)] = 12 (using log(p) = log(q) giving p=q for base being the same) x^2+4x-12 = 0 (adding -12 to both the sides) x^2+(6x-2x)-12= 0 [writing the mid term as the sum of two terms such that their product gives the product of the square term and the constant term and here 4x = (6x-2x) with (6x)X(-2x) = -12x^2 =(x^2)X(-12) ] (x^2+6x)-2x-12=0 (by additive associativity) x(x+6)-2(x+6) = 0 xp-2p = 0 p(x-2) = 0 where p = (x+6) (x+6)(x-2) = 0 x+6 = 0 gives x = -6 and as log(of a negative number is not defined we CANNOT HAVE x = -6 (x-2) = 0 gives x=2 And putting x=2 in (1) we have LHS = logx-log3 = log2-log3 = log(2/3) And RHS = log4-log(x+4)= log4 - log6 = log(4/6) = log(2/3) = LHS Therefore x = 2 is correct