Question 852545
I'll take a shot at it
{{{y=2e^(2x)}}}
{{{y=e^x -1}}}
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By substitution:
{{{ 2*e^(2x) = e^x - 1 }}}
{{{ 2*(e^x)^2 = e^x - 1 }}}
{{{ 2*(e^x)^2 - e^x + 1 = 0 }}}
Make the substitution:
{{{ z = e^x }}}
{{{ 2z^2 - z + 1 = 0 }}}
Use quadratic formula
{{{ z = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}} 
{{{ a = 2 }}}
{{{ b = -1 }}}
{{{ c= 1 }}}
{{{ z = ( -(-1) +- sqrt( (-1)^2 - 4*2*1 )) / (2*2) }}} 
{{{ z = ( 1 +- sqrt( -7 )) / 4 }}} 
{{{ z = ( 1 + 7i ) / 4 }}}
and
{{{ z = ( 1 - 7i ) / 4 }}}
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substituting again:
{{{ e^x = ( 1 + 7i ) / 4 }}}
{{{ x = ln( 1/4 + (7/4)*i ) }}}
and
{{{ x = ln( 1/4 - (7/4)*i ) }}}
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that's as far as I can go