SOLUTION: I am having trouble figuring out how to solve this problem. Find all real solutions to the equation 40-106e^x+101e^2x-41e^3x+6e^4x=0.

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Question 1059243: I am having trouble figuring out how to solve this problem.
Find all real solutions to the equation 40-106e^x+101e^2x-41e^3x+6e^4x=0.

Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
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I am having trouble figuring out how to solve this problem.
Find all real solutions to the equation 40-106e^x+101e^2x-41e^3x+6e^4x=0.
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Let us introduce new variable z = e%5Ex, Then the equation will take a form

p(z) = 6z^4 - 41z^3 +101z^2 - 106z + 40 = 0.


One root is z=1. You can check it making direct calculation.
So, the polynomial has the divisor (z-1).


The quotient p%28z%29%2F%28z-1%29 is q(z) = 6z%5E3+-+35z%5E2+%2B+66z+-+40 (long division).


The root of the cubic polynomial is z=2. You can check it making direct calculation.
So, the polynomial has the divisor (z-2).


The quotient q%28z%29%2F%28z-2%29 is r(z) = 6z%5E2+-+23x+%2B+20 (long division).


r(z) has the roots 2.5 and 4%2F3 (quadratic formula).


So, the polynomial p(z) has the roots  1, 2, 2.5 and 4%2F3.


Since z = e%5Ex%29, the original polynomial has the roots 


x = ln(1) = 0, x = ln(2), x = ln(2.5) and x = ln((3/4)).

Solved.