SOLUTION: Log1/3(x2+x)-log1/3(x2-x)=-1
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Question 725775: Log1/3(x2+x)-log1/3(x2-x)=-1
Found 2 solutions by solver91311, lwsshak3:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The difference of the logs is the log of the quotient.
\ -\ \log_{\small{\frac{1}{3}}\LARGE}\left(x^2\ -\ x\right)\ =\ -1)
\ =\ -1)
Use the definition of the log function:
 \ \ \Rightarrow\ \ b^y\ =\ x)
to write:
^{-1}\ =\ \left(\frac{x^2\ +\ x}{x^2\ -\ x}\right))
Then solve for 
but note that the root of the rational expression is NOT in the domain either of one of the original log functions nor in the domain of the resulting rational function. The solution set is the empty set.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Log1/3(x2+x)-log1/3(x2-x)=-1
log1/3[(x^2+x)/(x^2-x)]=-1
convert to exponential form: base(1/3) raised to log of number(-1)=number((x^2+x)/(x^2-x)
(1/3)^-1=(x^2+x)/(x^2-x)
3=x(x+1)/x(x-1)
3x-3=x+1
2x=4x
x=2
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