document.write( "Question 499302: How to solve |3*e^(2x)-5*e^(x)|<2 ? \n" ); document.write( "
Algebra.Com's Answer #337555 by chessace(471)\"\" \"About 
You can put this solution on YOUR website!
Not too easily.
\n" ); document.write( "First note that e^2x = (e^x)^2 and use y = e^x
\n" ); document.write( "Then have |3y^2 - 5y| < 2
\n" ); document.write( "Then break into the usual 2 cases for absolute value:
\n" ); document.write( "3y^2 - 5y < 2 and 3y^2 - 5y > -2
\n" ); document.write( "Then find zeroes after sending over the +-2:
\n" ); document.write( "3y^2 - 5y -2 = 0 and 3y^2 - 5y + 2 = 0
\n" ); document.write( "These factor into
\n" ); document.write( "(3y+1)(y-2)=0 and (3y-2)(y-1)=0
\n" ); document.write( "With roots
\n" ); document.write( "-1/3, 2 and 2/3, 1
\n" ); document.write( "By considering large (+ and -) values for y, it's clear that the function
\n" ); document.write( "f(y) = 3y^2 - 5y as y increases in value,
\n" ); document.write( "is > 2 until = 2 at y = -1/3
\n" ); document.write( "[then passes through the origin]
\n" ); document.write( "is = -2 at y = 2/3 and is < -2 until
\n" ); document.write( "[it reaches its minimum of - 2.833 at y = 5/6]
\n" ); document.write( "is = -2 again at y = 1,
\n" ); document.write( "and inceases to 2 at y = 2, then is > 2.
\n" ); document.write( "So the valid intervals for y are
\n" ); document.write( "(-1/3, 2/3) and (1, 2).
\n" ); document.write( "Now we need to relate this to x, where y = e^x.
\n" ); document.write( "So take natural logs of all 4 endpoints.
\n" ); document.write( "The -1/3 is artificial, e^x is never < 0.
\n" ); document.write( "Also e^0 = 1, so only need to look up 2 logs.
\n" ); document.write( "Final answer: x must belong to interval (-infinity, -0.405465) or
\n" ); document.write( "the interval (0 , 0.693147)
\n" ); document.write( " \n" ); document.write( "
\n" );