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(52776)   (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 = , 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  is q(z) =  (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  is r(z) =  (long division).


r(z) has the roots 2.5 and  (quadratic formula).


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


Since z = , the original polynomial has the roots 


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

Solved.



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