SOLUTION: solve the equation. log 1/5(x^2+x)- log 1/5(x^2-x)=-1

Question 970639: solve the equation.
log 1/5(x^2+x)- log 1/5(x^2-x)=-1
Found 2 solutions by khai, MathTherapy:
Answer by khai(18) (Show Source): You can put this solution on YOUR website!
first of all, is that a natural log (base 10) ? the way you write makes me confuse... if yes,
log (1/5) [(x^2+x)-(x^2-x)] = -1
log (1/5) [x^4-x^2] = -1
10^(-1) = (1/5)( x^4-x^2)
1/10 = (1/5) (x^4-x^2)
1/2 = (x^4-x^2)
x = -1.169 (2 multiplicity)
x = 1.169 (2 multiplicity)

Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

solve the equation.
log 1/5(x^2+x)- log 1/5(x^2-x)=-1

If is the base, or if the setup is: , then
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