#log_5(x-1) + log_5(x-2)-log_5(x+6) =0#
Note: #log(1) = 0 # for any base.
#log_5(x-1)+log_5(x-2)-log_5(x+6)=log_5(1)#
#log_5(((x-1)(x-2))/(x+6)) = log_5(1)#
Note: #log_b(P) + log_b (Q) = log_b(PQ)# and #log_b(P) - log_b(Q) = log_b(P/Q)#
#((x-1)(x-2))/(x+6) = 1#
Now to solve the equation.
Start by cross multiplying.
#(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^2-2x-x+2#
#(x-1)(x-2)=x^2-3x+2#
Our problem now becomes
#x^2-3x+2 = x+6#
Subtracting #x+6# from both the sides we get.
#x^2-3x+2-x-6=0#
#x^2-4x-4=0#
Solving for #x# using the quadratic formula.
Quadratic formula
#x=(-b+-sqrt(b^2-4ac))/(2a)#
Here #a=1#, #b=-4# and #c=-4#
#x=(-(-4)+-sqrt((-4)^2-4(1)(-4)))/(2(1))#
#x=(4+-sqrt(16+16))/2#
#x=(4+-sqrt(16*2))/2#
#x=(4+-4sqrt(2))/2#
#x=(2+-2sqt(2))#
#x=2-2sqrt(2)# will give a negative number and won't satisfy the given equation. Therefore the solution is #x=2+2sqrt(2)#
