SOLUTION: Find the Solution set of the equation e^5x+4e^3x+4e^2x=-9-3e^x+5e^4x

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Question 1037153: Find the Solution set of the equation e^5x+4e^3x+4e^2x=-9-3e^x+5e^4x
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
e%5E%285x%29%2B4e%5E%283x%29%2B4e%5E%282x%29=-9-3e%5Ex%2B5e%5E%284x%29
<==> e%5E%285x%29+-+5e%5E%284x%29+%2B4e%5E%283x%29%2B4e%5E%282x%29+%2B+3e%5Ex+%2B9+=+0.
Now let w+=+e%5Ex.
Then the last equation is equivalent to
w%5E5+-+5w%5E4+%2B4w%5E3%2B4w%5E2+%2B+3w+%2B9+=+0.
By using the rational root theorem and trying out root candidates by synthetic division, you will find that --
w%5E5+-+5w%5E4+%2B4w%5E3%2B4w%5E2+%2B+3w+%2B9+=+%28w%2B1%29%28w-3%29%5E2%28w%5E2%2B1%29.
==> the zeros are w = -1, 3 (double), i, -i.
Therefore the solution set of the original equation is { ln(-1), ln3, ln3, lni, ln(-i)}
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.)