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
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