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

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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) About Me  (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) About Me  (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 1%2F5 is the base, or if the setup is: log+%281%2F5%2C+%28x%5E2+%2B+x%29%29+-+log+%281%2F5%2C+%28x%5E2+-+x%29%29+=+-+1, then highlight_green%28system%28x+=+%283%2F2%29_or%2Cx+=+1.5%29%29