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)\"\" \"About 
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\"e%5E%285x%29%2B4e%5E%283x%29%2B4e%5E%282x%29=-9-3e%5Ex%2B5e%5E%284x%29\"\r
\n" ); document.write( "\n" ); document.write( "<==> \"e%5E%285x%29+-+5e%5E%284x%29+%2B4e%5E%283x%29%2B4e%5E%282x%29+%2B+3e%5Ex+%2B9+=+0\".\r
\n" ); document.write( "\n" ); document.write( "Now let \"w+=+e%5Ex\".\r
\n" ); document.write( "\n" ); document.write( "Then the last equation is equivalent to \r
\n" ); document.write( "\n" ); document.write( "\"w%5E5+-+5w%5E4+%2B4w%5E3%2B4w%5E2+%2B+3w+%2B9+=+0\".\r
\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( "\"w%5E5+-+5w%5E4+%2B4w%5E3%2B4w%5E2+%2B+3w+%2B9+=+%28w%2B1%29%28w-3%29%5E2%28w%5E2%2B1%29\".\r
\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.)
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