Question 1000654
{{{ log(abs(x),(5x^2-1))>2 }}}
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(x) }}}
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{{{ log(abs(x),(5x^2-1)) > log( abs(x), x^2 ) }}}
( note that {{{ x^2 }}} will be positive even if {{{ x }}}
is negative, giving me {{{ +2 }}} )
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Now I can say:
{{{ 5x^2 - 1 > x^2 }}}
{{{ 4x^2 > 1 }}}
{{{ x^2 > 1/4 }}}
{{{ x >  1/2 }}}
and also:
{{{ x < -1/2 }}} 

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Also, you can't have a log to a positive base which
gives you a negative result, so:
{{{ 5x^2 >= 1 }}}
{{{ x^2 >= 1/5 }}}
{{{ x >= 1/sqrt(5) }}}
{{{ x <= -1/sqrt(5) }}}
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These conditions are satisfied by my answers
so {{{ x > 1/2 }}} and {{{ x < -1/2 }}} are correct
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Defintely get a 2nd opinion if you can.
I think I'm right, but not positive