Solve the equation
log[1/6](x^2+x)-log[1/6](x^2-x)=-2
raise 1/6 to the power
(x^2+x)/(x^2-x) = (1/6)^-2 = 36
(x+1)/(x-1) = 36
x = 37/35
To solve the equation log[1/6](x^2+x) - log[1/6](x^2-x) = -2, we can start by using the properties of logarithms.
First, let's recall the logarithmic property known as the quotient rule. According to this rule, log(base a) (b) - log(base a) (c) = log(base a) (b / c).
Using this property, we can rewrite the given equation as a single logarithm: log[1/6]((x^2+x) / (x^2-x)) = -2.
Next, we can simplify the expression inside the logarithm.
(x^2+x) / (x^2-x) can be further simplified by factoring out an x from both the numerator and denominator:
x(x+1) / x(x-1) = (x+1) / (x-1).
By simplifying, the equation becomes log[1/6]((x+1) / (x-1)) = -2.
To eliminate the logarithm, we can raise both sides of the equation to the power of the base (1/6):
(1/6)^(-2) = (x+1) / (x-1).
Simplifying the left-hand side, we get (36) = (x+1) / (x-1).
Now, we can multiply both sides of the equation by (x-1) to clear the fraction:
36(x-1) = x+1.
Expanding the left side of the equation, we have 36x - 36 = x + 1.
Combining like terms, we get 35x - 36 = 1.
Adding 36 to both sides of the equation, we have 35x = 37.
Finally, dividing both sides of the equation by 35, we find that x = 37/35 is the solution to the given equation.