First we have to find the domain of this equation. Since loglog functions are only defined for positive real numbers we have to solve inequalities:
(x+3)(x-2)>0(x+3)(x−2)>0 and 7x-11>07x−11>0
From first inequality we get: x<-3 vv x>2x<−3∨x>2
From the second we get: x> 1 4/7x>147
So we can write, that the domain is: D=(2;+oo)D=(2;+∞)
Now we can solve the equation:
log(x+3)(x-2)=log(7x-11)log(x+3)(x−2)=log(7x−11)
Since the base of logarythm is the same on both sides (none is written, so I suppose it is 1010), we can skip loglog signs and solve equation:
(x+3)(x-2)=7x-11(x+3)(x−2)=7x−11
x^2+x-6=7x-11x2+x−6=7x−11
x^2-6x+5=0x2−6x+5=0
Delta=6^2-4*1*5
Delta=36-20=16
sqrt(Delta)=4
x_1=(-b-sqrt(Delta))/(2a)
x_1=(6-4)/2=1
x_2=(-b+sqrt(Delta))/(2a)
x_2=(6+4)/2
x_2=5
Since x_1 !in D, the only solution of this equation is x_2=5
