document.write( "Question 1037153: Find the Solution set of the equation e^5x+4e^3x+4e^2x=-9-3e^x+5e^4x \n" ); document.write( "
Algebra.Com's Answer #651880 by robertb(5830)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "\n" ); document.write( "<==> \n" ); document.write( "\n" ); document.write( "Now let \n" ); document.write( "\n" ); document.write( "Then the last equation is equivalent to \r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "By using the rational root theorem and trying out root candidates by synthetic division, you will find that --\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "==> the zeros are w = -1, 3 (double), i, -i.\r \n" ); document.write( "\n" ); document.write( "Therefore the solution set of the original equation is { ln(-1), ln3, ln3, lni, ln(-i)} \n" ); document.write( "If we are after REAL zeros only, then the solution to the original equation is x = ln3, which is a double root. ( ln(-1), lni, and ln(-i) all have values in the complex domain.) \n" ); document.write( " \n" ); document.write( " |