log_5(x-1) + log_5(x-2)-log_5(x+6) =0log5(x−1)+log5(x−2)−log5(x+6)=0
Note: log(1) = 0 log(1)=0 for any base.
log_5(x-1)+log_5(x-2)-log_5(x+6)=log_5(1)log5(x−1)+log5(x−2)−log5(x+6)=log5(1)
log_5(((x-1)(x-2))/(x+6)) = log_5(1)log5((x−1)(x−2)x+6)=log5(1)
Note: log_b(P) + log_b (Q) = log_b(PQ)logb(P)+logb(Q)=logb(PQ) and log_b(P) - log_b(Q) = log_b(P/Q)logb(P)−logb(Q)=logb(PQ)
((x-1)(x-2))/(x+6) = 1(x−1)(x−2)x+6=1
Now to solve the equation.
Start by cross multiplying.
(x-1)(x-2)=(x+6)(x−1)(x−2)=(x+6) As you can see this removes the denominator and makes it easier for us to solve.
Now simplify
(x-1)(x-2) = x(x-2)-1(x-2)(x−1)(x−2)=x(x−2)−1(x−2)
(x-1)(x-2)=x^2-2x-x+2(x−1)(x−2)=x2−2x−x+2
(x-1)(x-2)=x^2-3x+2(x−1)(x−2)=x2−3x+2
Our problem now becomes
x^2-3x+2 = x+6x2−3x+2=x+6
Subtracting x+6x+6 from both the sides we get.
x^2-3x+2-x-6=0x2−3x+2−x−6=0
x^2-4x-4=0x2−4x−4=0
Solving for xx using the quadratic formula.
Quadratic formula
x=(-b+-sqrt(b^2-4ac))/(2a)x=−b±√b2−4ac2a
Here a=1a=1, b=-4b=−4 and c=-4c=−4
x=(-(-4)+-sqrt((-4)^2-4(1)(-4)))/(2(1))x=−(−4)±√(−4)2−4(1)(−4)2(1)
x=(4+-sqrt(16+16))/2x=4±√16+162
x=(4+-sqrt(16*2))/2x=4±√16⋅22
x=(4+-4sqrt(2))/2x=4±4√22
x=(2+-2sqt(2))x=(2±2sqt(2))
x=2-2sqrt(2)x=2−2√2 will give a negative number and won't satisfy the given equation. Therefore the solution is x=2+2sqrt(2)x=2+2√2
