SOLUTION: Please if someone could solve this inequality! Thank you for your time and help! {{{ log(abs(x),(5x^2-1))>2 }}}

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Question 1000654: Please if someone could solve this inequality!
Thank you for your time and help!
+log%28abs%28x%29%2C%285x%5E2-1%29%29%3E2+
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
log(abs(x))(5x^2-1)) > 2
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|x|^2 > 5x^2-1
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Graph the left and the right side on the same coordinate
systerm and determine when the left side is greater than the right.
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graph%28400%2C400%2C-10%2C10%2C-10%2C10%2C%28abs%28x%29%29%5E2%2C5x%5E2-1%29
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Ans::
Cheers,
Stan H.
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Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+log%28abs%28x%29%2C%285x%5E2-1%29%29%3E2+
By using the absolute value of +x+ for the base,
they are just making sure the base is positive.
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The trick here, I think, is to express +2+ as a
log to the base +abs%28x%29+
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+log%28abs%28x%29%2C%285x%5E2-1%29%29+%3E+log%28+abs%28x%29%2C+x%5E2+%29+
( note that +x%5E2+ will be positive even if +x+
is negative, giving me +%2B2+ )
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Now I can say:
+5x%5E2+-+1+%3E+x%5E2+
+4x%5E2+%3E+1+
+x%5E2+%3E+1%2F4+
+x+%3E++1%2F2+
and also:
+x+%3C+-1%2F2+
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Also, you can't have a log to a positive base which
gives you a negative result, so:
+5x%5E2+%3E=+1+
+x%5E2+%3E=+1%2F5+
+x+%3E=+1%2Fsqrt%285%29+
+x+%3C=+-1%2Fsqrt%285%29+
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These conditions are satisfied by my answers
so +x+%3E+1%2F2+ and +x+%3C+-1%2F2+ are correct
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Defintely get a 2nd opinion if you can.
I think I'm right, but not positive